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    gere mechanics of materials solution manual

    You can change your ad preferences anytime. Full clear download at:Problem 2.2-1Full clear download at:Now customize the name of a clipboard to store your clips. For full document please download An Instructor’s Solutions Manual to AccompanyContentsAnswers to Problems 944Normal Stress and Strain P1. Problem 1.2-1 A hollow circular post ABC (see figure) supports a load. A tAB dAB P2. B dBC tBC C. Solution 1.2-1 PART (b)Tension, Compression, and ShearT AeHand brake pivot AUniform hand brake pressurePivot points anchored to frameRB Apad. Page 4. T AcableRB ApadSolution 1.2-4Problem 1.2-5. Normal Stress and StrainTension, Compression, and Shear. Problem 1.2-6. A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). Calculate the tensile stress st in the cable. Solution 1.2-6. Car on inclined track TENSILE STRESS IN THE CABLEThe shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are idealized as shown in the second part of the figure. Note that the base of the wall and both ends of the shores are assumed to be pinned.T1 A wireTension, Compression, and Shear. Solution 1.2-8T Ae d LcProblem 1.2-10 Solve the preceding problem if the. CaCrate. TruckTension, Compression, and ShearL T H v. Problem ?1.2-11 An L-shaped reinforced concrete slab. F Coordinates of D in ftMultiply unit forces by WTension, Compression, and Shear. A round bar ACB of length 2L (see figure) rotates about an axis through the midpoint C with constant angular speed v (radians per second). The material of the bar has weight density g. (a) Derive a formula for the tensile stress sx in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress smax? Solution 1.2-12. Problem 1.2-13 Two gondolas on a ski lift are locked in the position shown in the figure while repairs are being made elsewhere.TBC AeTension, Compression, and ShearProblem 1.

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    To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. Related Papers Chapter 09 By Andres Hernandez Nonprismatic Beams By Randall Monasterio READ PAPER Download pdf. Solutions Manuals are available for thousands of the most popular college and high school textbooks in subjects such as Math, Science ( Physics, Chemistry, Biology ), Engineering ( Mechanical, Electrical, Civil ), Business and more. Understanding Mechanics of Materials homework has never been easier than with Chegg Study. Unlike static PDF Mechanics of Materials solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. You can check your reasoning as you tackle a problem using our interactive solutions viewer. Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available. Hit a particularly tricky question. Bookmark it to easily review again before an exam. The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price. Why buy extra books when you can get all the homework help you need in one place? Just post a question you need help with, and one of our experts will provide a custom solution. You can also find solutions immediately by searching the millions of fully answered study questions in our archive. Asking a study question in a snap - just take a pic. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.If you wish to opt out, please close your SlideShare account. Learn more.

    2% offset. Is the material ductile or brittle?Mechanical Properties of Materials. Stress (MPa). StrainTensile test of a plastic. Using the stress-strain data given in the problem statement, plot the stress-strain curve. Modulus of elasticity (slope)Problem 1.3-7 The data shown in the accompanying table were obtained from a tensile test of high-strength steel. TENSILE-TEST DATA FOR PROBLEM 1.3-7 Load (lb)Solution 1.3-7. Tensile test of high-strength steelPage 20The yield stress of the steel is 42 ksi and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 30 103 ksi. The bar is loaded axially until it elongates 0.20 in., and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.). Solution 1.4-1. Problem 1.4-2 A bar of length 2.0 m is made of a structural steel having the stress-strain diagram shown in the figure. The yield stress of the steel is 250 MPa and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 200 GPa. The bar is loaded axially until it elongates 6.5 mm, and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.)Page 22. Final length of bar is 4.0 mm greater than its original length. Solution 1.4-3 RESIDUAL STRAINA circular bar of magnesium alloy is 750 mm long. The stressstrain diagram for the material is shown in the figure. The bar is loaded in tension to an elongation of 6.0 mm, and then the load is removed. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.). Elasticity, Plasticity, and CreepDetermine the elongation of the wire due to the forces P. If the forces are removed, what is the permanent set of the bar. If the forces are applied again, what is the proportional limit?

    2-14 A crane boom of mass 450 kg with itsCrSolution 1.2-14 DataMechanical Properties of Materials Problem 1.3-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon. (a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi? (b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea water from Table H-1, Appendix H.). Solution 1.3-1Page 16. Tension, Compression, and Shear. Hanging wire of length L (b) WIRE HANGING IN SEA WATERSolution 1.3-3Percent reduction in areaMaterial. L1 (in.)Brittle or Ductile?Problem 1.3-4 The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application.Note that the ratio has units of length. Solution 1.3-4. Strength-to-weight ratioTitanium has a high strength-to-weight ratio, which is why it is used in space vehicles and high-performance airplanes. Aluminum is higher than steel, which makes it desirable for commercial aircraft. Some woods are also higher than steel, and nylon is about the same as steel.Tension, Compression, and Shear. Problem 1.3-5 A symmetrical framework consisting of three pin-connected bars is loaded by a force P (see figure). The axial strain in the middle bar is measured as 0.0713. Determine the tensile stress in the outer bars if they are constructed of aluminum alloy having the stress-strain diagram shown in Fig. 1-13. (Express the stress in USCS units.)Aluminum alloy. Use stress-strain diagram of Figure 1-13Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.

    6-8 An elastomeric bearing pad consisting of two steel plates bonded to a chloroprene elastomer (an artificial rubber) is subjected to a shear force V during a static loading test (see figure). When the force V equals 12 kN, the top plate is found to have displaced laterally 8.0 mm with respect to the bottom plate. What is the shear modulus of elasticity G of the chloroprene? P A snV abTension, Compression, and Shear. Problem 1.6-9 A joint between two concrete slabs A and B is filled with a flexible epoxy that bonds securely to the concrete (see figure). Solution 1.6-9Shear Stress and StrainThe diameter of the pin is 0.5 in. and the thickness of the shackle is 0.25 in. The buoy has a diameter of 60 in. Solution 1.6-11Tension, Compression, and ShearBecause arm B straddles arm A, the pin is in double shear. Solution 1.6-12. Clamp supporting a load PThe overturning effect of the bikes on the rack is resisted by a force couple Fh at BC. (a) (b) (c) (d)Tension, Compression, and Shear. Bike loadsC Fixed support at AW1 Ax APin at C 2.125 in. D Bolt at BB res A bB. Bx A sC. Cx AbCYou might wish to examine a bicycle chain and observe its construction. Bicycle chainSolution 1.6-14Shear Stress and StrainTension, Compression, and ShearProblem 1.6-15 A shock mount constructed as shown in the figure is used to support a delicate instrument. Solution 1.6-15Member BC Support plate. Pin. By — 2. By — 2Tension, Compression, and Shear. Solution 1.6-16 (b) MAX. SHEAR STRESS IN PIN AT BPin at OWater pressure force on nozzle, f p. C (b). C Quick release fittings Garden hose (c) (a)Tension, Compression, and ShearOres AsAsStrut AB lies in a vertical plane.C HingeVECTOR rABFs A strutFsyFs AsTension, Compression, and Shear. Problem 1.6-19. The top portion of a pole saw used to trim small branches from trees is shown in the figure part (a).Collar. Saw blade. D aSolution 1.6-19 SOLVE ABOVE EQUATION FOR PSolution 1.7-1Tension, Compression, and Shear. Problem 1.

    Tension, Compression, and Shear. Solution 1.4-5. Wire stretched by forces PWhat is the largest compressive load Pmax that is permitted? Solution 1.5-1. Steel bar in compressionProblem 1.5-2 A round bar of 10 mm diameter is made of aluminum alloy 7075-T6 (see figure). Aluminum bar in tensionTension, Compression, and Shear. Steel tube d1 d2 Polyethylene bar. Solution 1.5-3 NORMAL STRAINTension, Compression, and Shear. Problem 1.5-6 A tensile test is peformed on a brass specimen 10 mm in diameter using a gage length of 50 mm (see figure). When the tensile load P reaches a value of 20 kN, the distance between the gage marks has increased by 0.122 mm. (a) What is the modulus of elasticity E of the brass? (b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio? Brass specimen in tensionP1 A dAB tAB P2 B Cap plate dBC tBC CProblem 1.5-8 A brass bar of length 2.25 m with a square cross section of 90 mm on each side is subjected to an axial tensile force of 1500 kN (see figure).Tension, Compression, and ShearP b. Floor slabAngle bracket tF 2AbProblem 1.6-2. Truss members supporting a roof are connected to a 26-mm-thick gusset plate by a 22 mm diameter pin as shown in the figure and photo. The two end plates on the truss members are each 14 mm thick. Truss memberGusset plateSolution 1.6-2 NUMERICAL DATACross sectional area of pinTension, Compression, and Shear. Determine the following quantities. Problem 1.6-3Solution 1.6-3 (b) BEARING STRESS ON PIN FROM FLANGE PLATEV aB Bx. Typical rung. Shoe bolt at AAx A —y 2. A —y 2. Shoe b. Section at base. Solution 1.6-4 (a) SUPPORT REACTIONS. N mTension, Compression, and ShearUse dimensions show in the figure. Lower end of front brake cable D TBrake pads CIgnore the mass of the cables.Clevis and pin 2Solution 1.6-6 SOLUTION APPROACHTension, Compression, and ShearT1 tpNut t. Steel plateSolution 1.6-7 (c) AVE. SHEAR THROUGH NUTProblem 1.

    Use the force at the base of the spring as the redundant. Use numerical properties in the boxes below. (a) (b) (c) (d). What is the resulting force in the spring, Fk. What is the resulting force in the tube, Ft. What is the final length of the tube, Lf. Cap (assume rigid) Tube (d0, t, L, at, Et). Spring (k, L1 m2 IbP ? 11.10 kNProblem 8.4-5 Solve the preceding problem if the stress and dimensions are as follows: s1 ? 2450 psi, L ? 80 in., b ? 2.5 in., h ? 10 in., and d ? 2.5 in. Solution 8.4-5 L ? 80 in.Maximum principal stress at point D: RB ? 0I ? 208.333 in.4 y ? 2.5 in. My ?(? PL)y lb ? ? (0.96P) I I in.2. Q ? bd a txy ?Q ? 23.438 in.3. PRINCIPAL STRESSES s1 ? ?Applications of Plane Stress. Problem 8.4-6 A beam of wide-flange cross section (see figure) has the following dimensions: b ? 120 mm, t ? 10 mm, h ? 300 mm, and h1 ? 260 mm. The beam is simply supported with span length L ? 3.0 m. A concentrated load P ? 120 kN acts at the midpoint of the span. At a cross section located 1.0 m from the left-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis.Probs. 8.4-6 and 8.4-7. Solution 8.4-6. Simply supported beam sy ? 0Uniaxial stress: s1 ? 0 s2 ? ?82.7 MPa tmax ? 41.3 MPaL ? 3.0 mMaximum Stresses in Beams. Pure shear: s1 ? 23.2 MPa, s2 ? ?23.2 MPa K tmax ? 23.2 MPaSolution 8.4-7. Simply supported beam. L ? 10 ft ? 120 in.Uniaxial stress: s1 ? 15,870 psi, d s2 ? 0 tmax ? 7930 psiQ ? baVQ (12,000 lb)(21.094 in.3) ?? It (285.89 in.4)(0.5 in.)My (756,000 lb-in.)( ?5.25 in.) ?? I 285.89 in.4Applications of Plane StressVQ (12,000 lb)(27.984 in.3) ?? It (285.89 in.4)(0.5 in.)Problem 8.4-8 A W 200 ? 41.7 wide-flange beam (see Table E-1(b). Appendix E) is simply supported with a span length of 2.5 m (see figure). The beam supports a concentrated load of 100 kN at 0.9 m from support B. At a cross section located 0.

    7 m from the left-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis.Solution 8.4-8 RB ? 100 kN a. RA ? 100 kN ? RB. RB ? 64 kN (upward). RA ? 36 kNAt the point D M ? RA(0.7 m) V ? RA. V ? 36 kNUniaxial stress: s1.MaQ ? 1.892 ? 105 mm3VQ I twVQ I twPure shear: s1 ? ? txy ?Q ? 2.19 ? 105 mm3Solution 8.4-9 RB ?RB ? 2.5 k (upward) (downward). V ? ?2.5 kUniaxial stress: s1.MaPage 672. Applications of Plane Stress. VQ I twQ ? 8.502 in3VQ I twPure shear:Determine the principal stresses s1 and s2 and the maximum shear stress tmax at points A and B in the web of the beam near the support. B 6.5 kNC 160 mmSolution 8.4-10 P ? 6.5 kN. L ? 2.5 m. A ? 8 ? 10 mm 3. Location of centroid CA ? 2(160 mm)(25 mm)From Eq. (12-7b) in Chapter 12:PH ? ?3.25 kN. Mc ? PH c1. Pv ? P sin (60 deg)V ? Pv. V ? 5.629 kN. M ? ?Mc ? Pv L. Stress at point A (bottom of web) sx ?Uniaxial stress: s1 ? sy s1 ? 0Stress at point B (top of web) sx. M1c1 ? t2 N0 ? A Iz. Q ? bt ac1 ? txy ? ? s1 ? s2 ?VQ Iz tQ ? 1.85 ? 105 mm3Applications of Plane Stress. L M at — 2The height of the beam is h ? 6 in. and the width is b ? 2.5 in. Plot graphs of the principal stresses s1 and s2 and the maximum shear stress tmax, showing how they vary over the height of the beam at cross section mn, which is located 24 in.Probs. 8.4-11 and 8.4-12. Solution 8.4-11 2 ? 103Prin. stresses (psi)RA ? 9.032 kRB ? ?9.032 kAt section m-n V ? RAI ? 45 in4. RB ? ?RA M ? RAc. VQ(y) IbSolution 8.4-12 sx( y) ? Prin. stresses (MPa)Q( y) ? b a txy( y) ?VQ( y) IbMy IRB ? ?86.667 kNAt section m-n M ? RA cCombined Loadings y0. The problems for Section 8.5 are to be solved assuming that the structures behave linearly elastically and that the stresses caused by two or more loads may be superimposed to obtain the resultant stresses acting at a point.

    7-2 A torque T0 is transmitted between two flanged shafts by means of ten 20-mm bolts (see figure and photo). If the allowable shear stress in the bolts is 90 MPa, what is the maximum permissible torque? (Disregard friction between the flanges.)Shafts with flangesIf the allowable shear stress in the fiberglass is 300 psi, and the allowable bearing pressure between the washer and the fiberglass is 550 psi, what is the allowable load Pallow on the tie-down?The corresponding yield and ultimate values in shear for the pin are 80 MPa and 140 MPa, respectively. Finally, the yield and ultimate values in bearing between the pins and the tubes are 260 MPa and 450 MPa, respectively. Assume that the factors of safety with respect to yield stress and ultimate stress are 4 and 5, respectively.Which is the controlling value of P?Tension, Compression, and Shear. Solution 1.7-4 Yield and ultimate stresses (all in MPa)Problem 1.7-5 A steel pad supporting heavy machinery rests on four short, hollow, cast iron piers (see figure). The ultimate strength of the cast iron in compression is 50 ksi.Cast iron piers in compression. Four piersF C MhByEyelet. Pin supportTension, Compression, and Shear. Fa2Cables attached to the lifeboat pass over the pulleys and wind around winches that raise and lower the lifeboat. The allowable tensile force in each cable is 1800 lb, and the allowable shear stress in the pins is 4000 psi. If the lifeboat weighs 1500 lb, what is the maximum weight that should be carried in the lifeboat? Solution 1.7-7. Lifeboat supported by four cablesShear in the pins governs.Shear stress in the pins is 50 MPa; bearing stress between the pin and the pulley is 110 MPa. aPinCage at BTension, Compression, and Shear. Solution 1.7-8 OR check bearing stressWmax2Problem 1.7-9 A ship’s spar is attached at the base of a mast by a pin connection (see figure). Determine the allowable compressive force Pallow in the spar. Mast P. Spar Connecting plateProblem 1.

    7-10 What is the maximum possible value of the clamping force C in the jaws of the pliers shown in the figure if the ultimate shear stress in the 5-mm diameter pin is 340 MPa. What is the maximum permissible value of the applied load P if a factor of safety of 3.0 with respect to failure of the pin is to be maintained? P yRxC PinTension, Compression, and Shear. Solution 1.7-10 NUMERICAL DATAAsFind PmaxProblem 1.7-11 A metal bar AB of weight W is suspended by a system of steelDetermine the maximum permissible weight Wmax for a factor of safety of 1.9 with respect to yielding.Problem 1.7-12 A plane truss is subjected to loads 2P and P at joints B and C, respectively, as shown in the figure part (a). The ultimate stresses in shear and bearing for the rivet steel are 190 MPa and 550 MPa, respectively. Determine the allowable load Pallow if a safety factor of 2.5 is desired with respect to the ultimate load that can be carried. (Consider tension in the bars, shear in the rivets, bearing between the rivets and the bars, and also bearing between the rivets and the gusset plate. Disregard friction between the plates and the weight of the truss itself.)C aP (c) Gusset plateTension, Compression, and ShearPallowAxially Loaded Members. Solution 2.5-23The figure shows a section through the tube, cap and springFlexibilitiesProblem 2.5-24 Prestressed concrete beams are sometimes manufactured in the following manner. High-strength steel wires are stretched by a jacking mechanism that applies a force Q, as represented schematically in part (a) of the figure. Concrete is then poured around the wires to form a beam, as shown in part (b). Thus, the beam is left in a prestressed condition, with the wires in tension and the concrete in compression.Thermal Effects. STRESSES (Eq. 1)Axially Loaded Members. Problem 2.5-25 A polyethylene tube (length L) has a cap which is held in place by a spring (with undeformed length L1 L). Ignore deformations of the cap and base.

    Consider both in-plane and out-of-plane shear stresses unless otherwise specified.The arms BC and CD have lengths b1 ? 3.6 ft and b2 ? 2.2 ft, respectively. The outer and inner diameters of the bracket are d2 ? 7.5 in. and d1 ? 6.8 in. A vertical load P ? 1400 lb acts at point D. Determine the maximum tensile, compressive, and shear stresses in the vertical arm. AApplications of Plane Stress. Solution 8.5-1 b1 ? 3.6 ftP ? 1400 lbA ? 7.862 in.2I ? 50.36 in.4MaEach arm is offset by the distance b ? 180 mm from the line of action of the weight force W. The allowable stresses in the arms are 100 MPa in tension and 50 MPa in shear. If the loaded gondola weighs 12 kN, what is the minimum diameter d of the arms?W (a)Gondola on a ski liftFind dmin 2Combined LoadingsSolve numerically: d ? 0.04845 m ? dmin ? 48.4 mmProblem 8.5-3 The hollow drill pipe for an oil well (see figure) is 6.2 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 62 k and the torque (due to drilling) is 185 k-in. Determine the maximum tensile, compressive, and shear stresses in the drill pipe.Applications of Plane Stress. Solution 8.5-3 P ? compressive force T ? TorqueTaIp. PRINCIPAL STRESSES s1 ?A ? 12.841 in.2 Ip ? 97.16 in.STRESSES AT THE OUTER SURFACE st ? s1. MAXIMUM TENSILE STRESS st ? 3963 psiNOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress.The shaft is hollow (outer diameter d2 ? 300 mm and inner diameter d1 ? 250 mm) and delivers 1800 kW at 4.0 Hz. If the compressive force P ? 540 kN, what are the maximum tensile, compressive, and shear stresses in the shaft?T ? torgue. P0 2p fP ? 540 kNP0 ? 1800 kW. P0 2p fA ? 2.16 ? 104 mm2. Ip ?IpIp ? 4.117 ? 108 mm4Problem 8.5-5 A segment of a generator shaft of hollow circular cross section is subjected to a torque T ? 240 k-in. (see figure).

    The outer and inner diameters of the shaft are 8.0 in. and 6.25 in., respectively. What is the maximum permissible compressive load P that can be applied to the shaft if the allowable in-plane shear stress is tallow ? 6250 psi?Applications of Plane Stress. Solution 8.5-5 P ? compressive force d2.T ? Torque d1 ? inner diameterA ? 19.586 in.2 Ip ?Ip. MAXIMUM IN-PLANESHEAR STRESS aP ? 194.2 kProblem 8.5-6 A cylindrical tank subjected to internal pressure p is simultaneously compressed by an axial force F ? 72 kN (see figure). The cylinder has diameter d ? 100 mm and wall thickness t ? 4 mm. Calculate the maximum allowable internal pressure pmax based upon an allowable shear stress in the wall of the tank of 60 MPa. Solution 8.5-6F ? 72 kN. Units: s1 ? MPaCombined LoadingsSolving, p2 ? 9.60 MPa Case 3: tmax ?Problem 8.5-7 A cylindrical tank having diameter d ? 2.5 in. is subjected to internal gas pressure p ? 600 psi and an external tensile load T ? 1000 lb (see figure). Determine the minimum thickness t of the wall of the tank based upon an allowable shear stress of 3000 psi. Solution 8.5-7T ? 1000 lbApplications of Plane Stress. LONGITUDINAL STRESS (TENSION) s2 ?Solving, t 1 ? 0.0413 in. OUT-OF-PLANE SHEAR STRESSES Case 2: tmax ?Solving, t2 ? 0.125 in. Case 3: tmax ?Solving, t3 ? 0.0837 in. CASE 2, OUT-OF-PLANE SHEAR STRESS GOVERNS tmin ? 0.125 in.Problem 8.5-8 The torsional pendulum shown in the figure consists of a horizontal circular disk of mass M ? 60 kg suspended by a vertical steel wire (G ? 80 GPa) of length L ? 2 m and diameter d ? 4 mm. Calculate the maximum permissible angle of rotation fmax of the disk (that is, the maximum amplitude of torsional vibrations) so that the stresses in the wire do not exceed 100 MPa in tension or 50 MPa in shear.Combined LoadingsL ? 2.0 mM ? 60 kg. G ? 80 GPaNote that s1 is positive and s2 is negative. Therefore, the maximum in-plane shear stress is greater than the maximum out-of-plane shear stress.

    MAXIMUM ANGLE OF ROTATION BASED ON TENSILE STRESS s1.GIp fmax LGIpfmax LTr IpUnits: t ? MPaApplications of Plane Stress. Problem 8.5-9 Determine the maximum tensile, compressive, and shear stresses at points A and B on the bicycle pedal crank shown in the figure. The pedal and crank are in a horizontal plane and points A and B are located on the top of the crank. The load P ? 160 lb acts in the vertical direction and the distances (in the horizontal plane) between the line of action of the load and points A and B are b1 ? 5.0 in., b2 ? 2.5 in. and b3 ? 1.0 in. Assume that the crank has a solid circular cross section with diameter d ? 0.6 in.Solution 8.5-9 P ? 160 lbTA ? Pb2. Moment. MA ? Pb1. Shear Force:STRESS RESULTANTS at point B:TB ? P(b1 ? b3)MB ? P(b2 ? b3) VB ? PMAXIMUM COMPRESSIVE STRESS sc ? ?2226 psiCombined LoadingsApplications of Plane Stress. Solution 8.5-10. Cylindrical pressure vessel s1 ? 56.4 MPaSTRESSES IN THE WALL OF THE VESSEL pr ? 25 MPa sx ? 2t Tr txy ? ? Ip T 2pr 2tUnits: txy ? MPaCombined LoadingsThe bracket has a hollow rectangular cross section with thickness t ? 0.125 in. and outer dimensions b ? 2.0 in. and h ? 3.5 in. The centerline lengths of the arms are b1 ? 20 in. and b2 ? 30 in. Considering only the load P, calculate the maximum tensile stress st, maximum compressive stress sc, and maximum shear stress tmax at point A, which is located on the top of the bracket at the support.L-shaped bracket b1 ? 20 in.STRESSES AT POINT A ON THE TOP OF THE BRACKET t?STRESS RESULTANTS AT THE SUPPORT Torque: Moment. T ? Pb2 ? (150 lb)(30 in.) ? 4500 lb-in. M ? Pb1 ? (150 lb)(20 in.) ? 3000 lb-in. Shear force. V ? P ? 150 lb. PROPERTIES OF THE CROSS SECTION For torsion: Am ? (b ? t)(h ? t) ? (1.875 in.)(3.375 in.) ? 6.3281 in.2 For bending: c ? I? ?Applications of Plane Stress. PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS s1, 2 ?MAXIMUM TENSILE STRESS: st ?

    4320 psiThe bar has centerline radius R and weight q per unit of length (total weight of the bar equals pqR). The cross section of the bar is circular with diameter d. Obtain formulas for the maximum tensile stress st, maximum compressive stress sc, and maximum in-plane shear stress tmax at the top of the bar at the support due to the weight of the bar. Solution 8.5-12R ? radius of barBending moment at B: MB. Wc ? 2qR2 Torque at B: TB. WR ? pqR2 (Shear force at B produces no shear stress at the top of the bar.)MAXIMUM TENSILE STRESS st ? s1 ?PRINCIPAL STRESSES: s1, 2 ?Problem 8.5-13 An arm ABC lying in a horizontal plane and supported at A (see figure) is made of two identical solid steel bars AB and BC welded together at a right angle. Each bar is 20 in. long. Knowing that the maximum tensile stress (principal stress) at the top of the bar at support A due solely to the weights of the bars is 932 psi, determine the diameter d of the bars.Horizontal arm ABC L.Applications of Plane StressSOLVE FOR dSTRESS ELEMENT ON TOP OF THE BAR AT A s1.Units: sx ? MPa, P ? newtons sy ?CROSS SECTION A ? 2prt ? 2p(50 mm)(3.0 mm) ? 942.48 mm2IP ? 2pr3t ? 2p(50 mm)3(3.0 mm)Problem 8.5-15 A post having a hollow circular cross section supportsApplications of Plane Stress. Solution 8.5-15 (a) MAXIMUM. P ? 240 lb b ? 5 ftI ? 67.507 in.4. Vx ? P cos(30 deg) Vy ? ?P sin(30deg)REACTIONS AT THE SUPPORT T ? ?P cos(30deg)bIp ? 135.014 in.4. M ? P cos(30deg)hA ? 7.059 in.2The stresses at point A are proportional to the load P. Based on tensile stress: Pallow sallow ? P smax. Pallow ?Pallow ? 847.01lb Based on shear stress: Pallow tallow.Mr2 I. Pallow ?Pallow ? 629 lbThe dimensions of the sign are 2.0 m ? 1.0 m, and its lower edge is 3.0 m above the base. Note that the center of gravity of the sign is 1.05 m from the axis of the pipe. The wind pressure against the sign is 1.5 kPa.

    Determine the maximum in-plane shear stresses due to the wind pressure on the sign at points A, B, and C, located on the outer surface at the base of the pipe. Rose’s Editing Co.PipeSolution 8.5-16 d2 ? 110 mm I?Ip ? 2IIp ? 7.933 ? 106 mm4STRESS RESULTANTS AT THE BASE M ? Ph. T ? PbV ? 3 kN. P ? 3 kNMd 2 2ITd 2 2IpApplications of Plane StressVQ Td2 ? 2 Ip I(2t)Problem 8.5-17 A sign is supported by a pole of hollow circularThe outer and inner diameters of the pole are 10.5 in. and 8.5 in., respectively. The pole is 42 ft high and weighs 4.0 k. The sign has dimensions 8 ft ? 3 ft and weighs 500 lb. Note that its center of gravity is 53.25 in. from the axis of the pole.Solution 8.5-17 PIPE:W1 ? 4000 lb SIGN. As ? (8)(3) ftArea. Height from the base to the center of gravity of the signTd2 2Ip. Md2 2IP ? pw AsP ? 840 lb. T ? PbMax. shear stressMax. compressive stress. M ? Ph. Nz ? W1 ? W2W2 ? 500 lb. NzApplications of Plane Stress. Problem 8.5-18 A horizontal bracket ABC consists of two perpendicular arms AB of length 0.5 m, and BC of length of 0.75 m. The bracket has a solid circular cross section with diameter equal to 65 mm.A 0.75 m. Frictionless sleeve embedded in supportB z0Solution 8.5-18 d ? 65 mmTorsional frictionless sleeve at support A (MZ). Vy ? ?M2 b1. Vy QIdA ? 3.318 ? 103 mm2. Pure ShearI ? 8.762 ? 105 mm4. STRESSES AT POINT p ON THE SIDE OF THE BRACKET. Ip ? 2 I Q? Ip ? 1.752 ? 106 mm4Q ? 2.289 ? 104 mm3. STRESS RESULTANTS AT SUPPORT A Nz ? 0. Axial force. My ? 0 Mx ? ? M2. Mx ? 0The outer radius is 12.0 in. and the wall thickness is 1.0 in. The loads are as follows: T ? 800 k-in., M ? 1000 k-in., and the internal pressure p ? 900 psi. Determine the maximum tensile stress st, maximum compressive stress sc, and maximum shear stress tmax in the wall of the cylinder.Solution 8.5-19. Cylindrical pressure vessal. Internal pressure: Bending moment:Torque. T ? 800 k-in. Outer radius:Wall thickness:Mean radius. Outer diameter:Inner diameter:Tr2 ? ?1002.7 psi Ip.


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    gerber shower faucet manual

    Sometimes, depending on the age of the toilet, it's better to replace them all at once than one at a time which in the long run is just more cost effective as your plumber comes once and all the parts are now new and should last longer. Some shipments may be delayed due to shutdowns beyond our control. We will do our best to notify customers with updates as we receive them. We will not be answering phones or checking emails during this time. Any communications sent will be answered in the order they were received upon our return. Thank you for your support. We will reopen January 2, 2020. Happy Holidays and Happy New Year! If you're planning on switching some of your fixtures out, we have you covered there as well. We offer Gerber faucet parts, Gerber toilet replacement parts and sinks. Each one is built to outlast many of your standard parts, and we want to be your retailer for Gerber faucet cartridges and Gerber plumbing supplies. If you're looking to give your bathroom or kitchen an elegant, fresh look, look no further than Gerber. It's an immediate way to update any room without breaking your budget. Gerber offers decorative bathroom sink faucets, tub-shower valves and accessories to match. All Gerber bathroom faucets are available in decorative finishes like brushed nickel.Gerber offers decorative kitchen faucets in amazing finishes like Chrome, Stainless Steel and Oil Rubbed Bronze.Gerber manufacturers high quality china products in several finish options for a perfect match. With more than 50,000 products online, always make Ferguson your first choice! Already a customer? Login All rights reserved.Refer to your local plumbing inspector or manufacturer for more details. If you want NextDay, we can save the other items for later. Order by, and we can deliver your NextDay items by. You won’t get NextDay delivery on this order because your cart contains item(s) that aren’t “NextDay eligible”.

    • gerber shower faucet parts, gerber shower faucet repair, gerber shower faucet repair kit, gerber shower faucet replacement, gerber shower faucet replacement parts, gerber shower faucet repair parts, gerber shower faucet manual, gerber shower faucet manual parts, gerber shower faucet manual troubleshooting, gerber shower faucet manual instructions, gerber shower faucet manual toilet.

    If there is no bonnet nut (in old models), then the temp limit device will be exposed. There is no need to remove any other part to adjust the temp limit device. New version is Blue and Red. Place the red cam on top. Position it so that both tapered end meets each other in one line as shown below. Note: This setting will allow for maximum hot water. Tighten with an adjustable wrench between 10 and Pull out from the body assembly. These indicate the amount of hot water which is allowed to flow. If it is too cold, turn the Temp Limit Device counter clockwise. Please upgrade your browser to improve your experience. Find Gerber cartridges, stems, handles, washers and rebuild kits to fix any Gerber fitting. Looking to update your old Gerber two or three handle shower to a more modern look. Be sure to check out our assortment of Gerber Brushed Nickel Trim components. Repair any Gerber two handle or three handle shower valve with genuine Gerber tub and shower parts.Flappers Fill Valves Flush Valves Tank Levers Gerber Toilet Parts Guide. Arrow must point up. 3.3 Secure retainer and tighten screws. 3.4 Replace plaster guard. Note: Up arrow on plaster guard and retention ring always need to be aligned, pointing up. Stops are open when screwdriver slot is horizontal. Stops are close when screwdriver slot is vertical. All while being surprisingly affordable. It was easy to install and we look forward to enjoying it for many years to come.” And for good reason. In the hottest housing markets, First, there’s the lingo: likes, shares, tweets, hashtags, and more Read more Agree Disagree. We also carry: Cartridges and Stems, Tub Waste Parts, Escutcheons, Face Plates, Handles, Handle Parts, Spray Heads, Faucets and more.Is the handle stuck or broken. It might be time to start replacing some parts. If the toilet is running it could be the fill valve, flush valve or flapper if it has one.

    com, we are committed to protecting your privacy. Your email address will never be sold or distributed to a third party for any reason. If you need immediate assistance, please contact Customer Care. Thank you Your feedback helps us make Walmart shopping better for millions of customers. OK Thank you! Your feedback helps us make Walmart shopping better for millions of customers. Sorry. We’re having technical issues, but we’ll be back in a flash. Done. Oct 2 - 3Our payment security system encrypts your information during transmission. We don’t share your credit card details with third-party sellers, and we don’t sell your information to others. To hide it, choose Ship in Amazon packaging at checkout.Please try again.Register a free business account Exclusive access to cleaning, safety, and health supplies. Create a free business account to purchase Show details Please try your search again later.You can edit your question or post anyway.In order to navigate out of this carousel, please use your heading shortcut key to navigate to the next or previous heading. To calculate the overall star rating and percentage breakdown by star, we don’t use a simple average. Instead, our system considers things like how recent a review is and if the reviewer bought the item on Amazon. It also analyzes reviews to verify trustworthiness. The parts feel solid and of good quality, like the ones I removed. No more leaks behind the faucet handles and no more hard-to-turn and squeaky handles. I watched three YouTube videos online because the kit does not come with instructions. The videos were helpful in illustrating the structure of a bathtub faucet and informing me of the additional plumber's tools I needed. I also used tools I already have like flathead and Phillips-head screwdrivers. I am not very strong, but with a little muscle and leverage, I was able to do this job on my own. Better yet, borrow or buy used plumber's tools so you only have to pay for the kit.

    In your cart, save the other item(s) for later in order to get NextDay delivery. Oops! There was a problem with saving your item(s) for later. You can go to cart and save for later there.Learn more This set features compression cartridges and a single function showerhead for easy switching from a classic faucet to a spray setting. The head of this chrome shower faucet is easy to clean and has soft-to-the-touch nozzles with pressure manifold and air injection technology for consistent water pressure and upkeep over time. About This Item We aim to show you accurate product information. Manufacturers,See our disclaimer Add a touch of simplicity to your bath with the Gerber Tub and Shower Faucet. The two valves control both the hot and cold water, allowing you to manually adjust to the desired temperature. This set features compression cartridges and a single function showerhead for easy switching from a classic faucet to a spray setting. Gerber Tub And Shower Faucet, 2-Valve: Model: 7-48-721 Manufacturer: Gerber Plumbing 2-valve water temperature controls Showerhead and Gerber tub faucet included Specifications Manufacturer Part Number G0748721 Model G0748721 Assembled Product Dimensions (L x W x H) 10.50 x 7.50 x 7.00 Inches Customer Reviews Write a review Be the first to review this item. Ask a question Ask a question If you would like to share feedback with us about pricing, delivery or other customer service issues, please contact customer service directly. So if you find a current lower price from an online retailer on an identical, in-stock product, tell us and we'll match it. See more details at Online Price Match. Related Pages: Delta Kitchen Faucets Tub and Shower Faucets Bathtub Faucets Shower Faucets Kraus Kitchen Faucets All Kitchen Faucets Bathroom Sink Faucets All Rights Reserved. To ensure we are able to help you as best we can, please include your reference number: Feedback Thank you for signing up. You will receive an email shortly at: Here at Walmart.

    If you were to buy all of these parts individually, they would be over 100 dollars, I know because I did this first and returned the old ones after buying this set. Gerber Plumbing Fixtures LLC.The Anatomy of Your Product GERBER Be sure that you’re buying genuine GERBER replacement parts. Description Chrome 1.Shower Assy. Not Sold 2.Shower Head. 49-199 3.Shower Arm. 49-111-G 4. See theexploded assembly for identification. Local Plumbing Code will specify the maximum temperaturesettings for the LIMIT STOP. 1. Gerber manufactures single handle, dual handle and triple handle faucet designs. The first is that the faucets receive a lot of wear-and-tear because of constant use. Secondly, hard water and mineral deposits combine to wear out the washers and seals within the faucet. Remove the screw in the center of the handle so that you can remove the handle. The screw may be concealed by a decorative cover that you will need to remove to access the screw. The cartridge should now be completely exposed. Use a pair of adjustable pliers to pull the cartridge out. Try to pull straight out so as to avoid damaging the fixture. Replace any parts that look old, cracked or worn and replace the fixture in the opposite order that you removed it. Turn the water back on at the main valve and test out faucet. If the leak is no longer present, you have repaired the shower faucet. If, however, the leak still persists, call a licensed plumber to determine what the problem is with your tub or shower faucet. Should your triumph be delayed, don't worry. We've compiled a few possible solutions for the most common post-installation issues with a new valve: little or no water coming out. This removes any sediment or debris that may have found its way into the supply lines or valve. If the valve isn't flushed, this debris can build up and create blockages, resulting in a reduction in pressure, little or no water, or only hot or cold water.

    So, I was very happy to save money on this faucet job!!!I find it takes too long to get in and out and I always forget something which adds significant time to the repair. But I had yet to buy any home repair items through Amazon and was a bit nervous. I read one of the reviews for this product that alerted me to a couple of tools I needed to buy and bought them at the same time. This kit fit perfectly and had everything I needed including the chrome sleeves, which are surprisingly hard to find. The parts themselves feel substantial to the touch and were identical to what I had previously only newer and shinier. The only thing I found is that when I tried to install the O-Ring washer for the shower handle, for some reason the handle wouldn't work (i.e. it didn't switch the water to the shower head). I left this off, and it works with no leaks.Note: Make Sure The the stem nuts are tight because they are shipped loose. I could have rebuilt the old ones but At this Price Don’t Bother.The original brand name and part number is a plus and even if an approximate year of the orginal can really aid in restoration projects like ours being a century old farmhouse. The Kisser 87-215 Gerber replacement is exactly what was needed for our restoration and works great!The last time my bathroom had been redone was in the early 90s so I knew it would be difficult finding the exact Gerber stem that would match my piping. I went to the local hardware stores (Home Depot and Lowes) and could not find the correct piece. After spending too much money I used internet shopping as a last option. This product was easy to install and it fit correctly with my old piping. I have had no problems in a couple weeks. I would recommend this product for anybody who has older Gerber pipes and cannot find the stems in stores. Good price as well.

    With thousands of parts covering over 100 brands our site has become a comprehensive guide that some manufacturers even use to see what fits their products. We ship to plumbers, maintenance workers, and direct to home owners. Some home owners do it themselves, while others get the parts to save time and money when the plumber comes. Use the menu or the links below to find the parts you need. Get Details. Model Number AB50-0142, AB50-6071, AB50-6072, AB555-1120, AB555-1130, Includes Seat, Package Quantity 10Model Number AB50-0142, AB50-6071, AB50-6072, AB555-1120, AB555-1130, Includes Seat, Package Quantity 10Model Number 07-43-431, Use With Water Temp. ColdModel Number 07-43-431, Use With Water Temp. ColdModel Number 07-43-431, Use With Water Temp. HotModel Number 07-43-431, Use With Water Temp. HotZoro Select Shower Rebuild Kit include characteristics like: Material: Brass.Zoro Select Shower Rebuild Kit include characteristics like: Material: Brass.Features: Pliers, Ruler, Scissors, Flathead Screwdriver, Phillips Head Screwdriver, Bottle Opener, File, Carbide Cutters, Saw, Can Opener.Features: Pliers, Ruler, Scissors, Flathead Screwdriver, Phillips Head Screwdriver, Bottle Opener, File, Carbide Cutters, Saw, Can Opener.Features: Deluxe Version of The Multi-Plier, Flick of The Wrist, One Handed Opening Pliers, Saf.T. Plus Locking Knife and Tools, RemGrit saw with universal saw coupler.Features: Deluxe Version of The Multi-Plier, Flick of The Wrist, One Handed Opening Pliers, Saf.T. Plus Locking Knife and Tools, RemGrit saw with universal saw coupler.We’ve got them at everyday low prices.We’ve got them at everyday low prices. We are ready to become your one-source solution. Call 1-800-345-3000 or email us to set your web login up for your existing account. Download our credit application to apply. Our digital help center is here to help you navigate Home Depot Pro's one-source solutions and get the job done right. Please Log In to use this helpful feature.

    Ships from manufacturer. Please note that additional delivery time may apply. This list updates each time you open it based on the criteria selected. A dripping faucet wastes both water and energy. Fixing your leaking faucet by repairing the stem will save water and keep you from buying an entire new faucet.For more information go to www.P65Warnings.ca.gov. Please call us. It may not display this or other websites correctly. You should upgrade or use an alternative browser. I have never done this type of repair before so I was hoping to get some clarification. I have attached several pictures including the type of Gerber handle (single), the current valve cartridge, and the two new parts I was provided. My question is whether the old parts just pull straight out and the new ones slide in or if there is more to it than that (such as removing additional pieces from the handle). I would appreciate any insight that can be provided. I have to shut off the water for the entire 6 unit building to make the change so I want to make sure I understand what I should be doing. Thanks! Mike I have never done this type of repair before so I was hoping to get some clarification. Thanks! Mike View attachment 22476 Click to expand. I am assuming that just screws off. Correct?Heard of so many first timers trying to change a shower cartridge without realizing that the water supply needs to be isolated.How about someone who does it at 2:00 a.m. in the morning in an apartment complex with about 18 water meters in many different locations and none of them labled. He was bailing the water out of the tub while I was trying to find the meter for his section of the building.How about someone who does it at 2:00 a.m. in the morning in an apartment complex with about 18 water meters in many different locations and none of them labled. He was bailing the water out of the tub while I was trying to find the meter for his section of the building. Click to expand.

    Remove the cartridge, clear out any obstructions from it and the valve, and turn on the water for a minute to flush out any debris that may still be in the valve where you can't see it.We do not recommend anyone try to deflect the water with just their bare hands, as hot water scalding could be a risk. After reinstalling the cartridge, normal flow should be restored. This should clear any blockages on the way to the shower head. While you have the head off the arm, you can also clean out any screens or flow restrictors. An incorrectly adjusted temperature stop can lead to water output that's too hot or too cold. The faucet installation instructions should have details on how to correctly set the temperature level for your shower. With a pressure-balance valve, if only one side has pressure (in this case, the cold side), the valve will do what it's designed to do and compensate for the lack of pressure on the other side. This results in either no water, or only a trickle making it through. Be sure that both hot and cold supplies are turned on! When shaken, the unit should rattle. If it's still stuck, you'll need to replace the spool. Opening this valve will allow that air to escape. If you have an automatic air vent, it could be plugged or malfunctioning. Unlike an appliance you'll get around to fixing at some point, when a home or building's most often used plumbing fixtures and accesories break or begin to fail, you and your customers need a swift solution. F.W. Webb is ready to help. They have been doing it for over 80 years and will continue to build their reputation for quality and steadfast loyalty to the trade professional. On top of matching every style and taste, the full-spectrum of plumbing products is also engineered with superb quality to last a lifetime. Simply complete this form and a member of our team will contact you promptly, typically within two business hours. Standard business hours are from 7:00 a.m. to 5:00 p.m.

    (ET), Monday through Friday, excluding major holidays. Send us a photo (optional). Click here to continue shopping Please do not confuse Factory Direct Supply with the original manufacturer. WARNING: Do Not Eat Our Products. California Propostion 65. Some of our products contain chemicals known to the State of California to cause cancer or other reproductive harm. If you are selling to anyone in California, we recommend that you label these items with the proper warning. You must have JavaScript enabled in your browser to utilize the functionality of this website. Ask Lester The Old Time Hardware Store Man Just answer the questions below and submit a photo. The more information you provide the greater the chance we will find the part you are looking for. If you select a particular brand, related Product List will appear. If you select a particular brand, related Product List will appear. Let us help you! Please Click Here to contact a knowledgeable PSC representative who will promptly reply to your request. Our Live Chat hours are:Add item to cart for lowest price.Manufacturer's warranty still applies.If you have problems accessing your account, please contact us at 1-888-757-4774 and we'll help you out. Join our mailing list to receive exclusive offers and coupons. Please review your friend's email address and try again. You can also view any product you subscribe to on the Back in Stock Subscriptions page in your Dashboard. Thank you for using Chadwell Supply, we look forward to serving you again in the near future. All the images are in thumbnail format, to see a You can get them at any plumbing supply. If not see the image with the handle remover. If you are going to replace the stems and handles any way just grab the handle with a large pair of Channel locks and push down and up and the handle will break off the stem.

    Apartment owner tried changing his shower cartridge in a condominium without shutting off the water. The guy is fighting against pumped hot water trying to force the cartridge back into the valve. Only the security downstairs had one of the 'T bar' keys to shut the water off in his unit.Apartment owner tried changing his shower cartridge in a condominium without shutting off the water. Only the security downstairs had one of the 'T bar' keys to shut the water off in his unit. Click to expand. My problem was that I had the key, but didn't know where the meter was or which one it was among many at each location. Fountains Those products can only be priced by contacting your local Coburn's location. We cannot guarantee the accuracy or completeness of the information, including price, images, spec, availability and services. Please review our Terms of use. Please contact your local Coburn's location if there is a requirement that the product comes from a specific manufacturer. From the intuitive open-and-close stops for easy maintenance, to easy installation and pressure-testing features, Treysta saves both time and money. Fully compatible with Danze by Gerber shower trim kits, Treysta helps you get the job done quickly and confidently giving you peace of mind backed by Gerber reliability. By Material Brass Bronze Chrome Copper Nickel Plastic Porcelain Stainless Steel more. Find low prices on Bathroom Faucets, Kitchen Faucets, Utility Sinks, and Delta Faucets. We currently do not have any products matching your search. Ads related to gerber shower handle replacement More information about Plumbing Supplies Best prices on Gerber shower handle replacement in Plumbing Supplies online. Visit Bizrate to find the best deals on top brands. Sales taxes and shipping costs are estimates; please check store for exact amounts. Product specifications are obtained from merchants or third parties.

    Although we make every effort to present accurate information, Bizrate is not responsible for inaccuracies. We encourage you to notify us of any discrepancies by clicking here. Please log in to your account to retrieve them.All exceptions will have a shipping fee listed on the product page. Exceptions include but are not limited to: This process usually takes 1 to 3 business days. All freight deliveries will require a signature for release. Please note any additional charges billed by the shipping company for services not requested by DecorPlanet.com (reconsignment, re-delivery after a missed appointment, storage fees, etc.) are the responsibility of the customer. Please note shipping fees, any potential duties, tariffs, customs or brokerage fees for shipments to Canada are the responsibility of the customer. Please note we reserve the right to split your order, which means you may receive shipments from various locations around the country. The delivery timeframe might vary depending on the product and your shipping address. When you bring the shipment into your home, carefully inspect each item for damages and report them to us as soon as possible. If the damages are noted on the proof of delivery, we will be able to accommodate cost-free replacements if the damages are reported to us within 14 days of the delivery receipt. This policy applies to both parcel and freight shipments, signature and not signature required. You are not required to purchase theThis Opulence single handle tub and shower faucet and we think you'll agree it's not an overstatement.These shower heads feature an air injection ball joint that adds bits of warm air to the shower's water flow. This process increases water velocity, minimizes pressure loss and provides complete and even coverage. They also come with a neo-pearl flow control aerator which provides constant water pressure and saves up to 20% on water consumption.

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    gere mechanics materials solution manual

    Assume that all forces act in the plane of the figure and that cable tension T 45 lbs. Also, what is the average compressive normal stress on the brake pad (A 0.625 in.2)? (b) For each braking system, what is the stress in the brake cable T (assume effective area of 0.00167 in.2)? (HINT: Because of symmetry, you only need to use the right half of each figure in your analysis.) 4 in. T D TDC TDE T TDE 4 in. TDC TDE E C TDCh 5 in. 4.25 in. RB A G Pivot points anchored to frame (a) Cantilever brakes Solution Apad 0.625 in.2 Acable 0.00167 in.2 (a) CANTILEVER FORCE RB PAD PRESSURE Statics: sum forces at D to get TDC T 2 a MA 0 RB(1) TDCh(3) TDCv(1) TDCh T 2 TDCh TDCv RB 90 lbs so RB 2T vs 4.25T for V brakes (below) HA B E RB 1 in. B F RB 2T T TDCv 2 in. Normal Stress and Strain A NOTE: xc yc are the same as expected due to symmetry about a diagonal 01Ch01.qxd 6 7:49 PM CHAPTER 1 Page 6 Tension, Compression, and Shear Problem A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track a steel cable (see figure). The cable has an effective area of 490 mm2, and the angle a of the incline is Calculate the tensile stress st in the cable. Solution Car on inclined track TENSILE STRESS IN THE CABLE DIAGRAM OF CAR W Weight of car T Tensile force in cable Angle of incline A Effective area of cable R1, R2 Wheel reactions (no friction force between wheels and rails) st T Wsin a A A SUBSTITUTE NUMERICAL VALUES: W 130 kN A 490 mm2 st (130 kN)(sin 490 mm2 133 MPa EQUILIBRIUM IN THE INCLINED DIRECTION 0 Q T W sin a 0 T W sin Problem Two steel wires support a moveable overhead camera weighing W 25 lb (see figure) used for viewing of field action at sporting events. At some instant, wire 1 is at on angle to the horizontal and wire 2 is at an angle Both wires have a diameter of 30 mils. (Wire diameters are often expressed in one mil equals 0.001 in.) Determine the tensile stresses and in the two wires.

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    Statically Indeterminate Beams Differential Equations of the Deflection Curve 795 Method of Superposition 809 Temperature Effects 839 Longitudinal Displacements at the Ends of Beams 843 11. Columns Idealized Buckling Models 845 Critical Loads of Columns with Pinned Supports 851 Columns with Other Support Conditions 863 Columns with Eccentric Axial Loads 871 The Secant Formula 880 Design Formulas for Columns 889 Aluminum Columns 903 v 00FM.qxd vi 8:49 PM Page vi CONTENTS 12. Review of Centroids and Moments of Inertia Centroids of Plane Ares 913 Centroids of Composite Areas 915 Moment of Inertia of Plane Areas 919 Theorem 923 Polar Moments of Inertia 927 Products of Inertia 929 Rotation of Axes 932 Principal Axes, Principal Points, and Principal Moments of Inertia 936 Answers to Problems 944 01Ch01.qxd 2 7:49 PM CHAPTER 1 Page 2 Tension, Compression, and Shear Part (c) P2 2260 dBC 2tBC P1 P2 ABC sAB P1 P2 2.744 sAB dBC tBC A dBC2 2 A dBC tBC 0.499 inches 4 P1 P2 a b p sAB Problem A force P of 70 N is applied a rider to the front hand brake of a bicycle (P is the resultant of an evenly distributed pressure). As the hand brake pivots at A, a tension T develops in the long brake cable (Ae 1.075 mm2) which elongates 0.214 mm. Find normal stress and strain in the brake cable. Brake cable, L 460 mm L 460 mm Ae 1.075 mm2 0.214 mm Statics: sum moments about A to get T 2P T Ae s 103.2 MPa d L 4.65 10 s 1.4 105 MPa NOTE: (E for cables is approx. 140 GPa) Hand brake pivot A 37.5 mm A P (Resultant of distributed pressure) mm 100 P 70 N T 50 Solution 4 P1 P2 b a p sAB 2 (dBC 2tBC)2 dBC2 4 P1 P2 b a p sAB mm Uniform hand brake pressure 01Ch01.qxd 7:49 PM Page 3 SECTION 1.2 3 Normal Stress and Strain Problem A bicycle rider would like to compare the effectiveness of cantilever hand brakes figure part versus V brakes part (a) Calculate the braking force RB at the wheel rims for each of the bicycle brake systems shown.

    dF Inertia force (centrifugal force) of element of mass dM g dF (dM)(jv2) g Av2jdj L g 2 gAv2 2 2 Av jdj (L ) 2g LD Lx g (a) TENSILE STRESS IN BAR AT DISTANCE x B Fx sx dF gv2 2 Fx (L x2) A 2g (b) MAXIMUM TENSILE STRESS Problem Two gondolas on a ski lift are locked in the position shown in the figure while repairs are being made elsewhere. The distance between support towers is L 100 ft. The length of each cable segment under gondola weights WB 450 lb and WC 650 lb are DAB 12 ft, DBC 70 ft, and DCD 20 ft. The cable sag at B is 3.9 ft and that at is 7.1 ft. The effective area of the cable is Ae 0.12 in2. (a) Find the tension force in each cable neglect the mass of the cable. (b) Find the average stress (s ) in each cable segment.A load P 20 kN is supported at point D. The crane boom lies in the plane. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. Related Papers Chapter 09 By Andres Hernandez Nonprismatic Beams By Randall Monasterio READ PAPER Download pdf. The 13-digit and 10-digit formats both work. Please try again.Please try again.Please try again. Then you can start reading Kindle books on your smartphone, tablet, or computer - no Kindle device required. Register a free business account To calculate the overall star rating and percentage breakdown by star, we don’t use a simple average. Instead, our system considers things like how recent a review is and if the reviewer bought the item on Amazon. It also analyzes reviews to verify trustworthiness. Solutions Manuals are available for thousands of the most popular college and high school textbooks in subjects such as Math, Science ( Physics, Chemistry, Biology ), Engineering ( Mechanical, Electrical, Civil ), Business and more. Understanding Mechanics of Materials homework has never been easier than with Chegg Study.

    The tailgate weighs WT 60 lb and is supported two cables (only one is shown in the figure). Each cable has an effective area Ae 0.017 in2. (a) Find the tensile force T and normal stress in each cable. (b) If each cable elongates 0.01 in. due to the weight of both the crate and the tailgate, what is the average strain in the cable. WC 150 lb H 12 in. dc 18 in. Ca ble Crate Truck Tail gate dT 14 in. L 16 in. WT 60 lb 01Ch01.qxd 7:50 PM Page 9 SECTION 1.2 Normal Stress and Strain Solution (a) T 2 Tv2 T h2 T 184.4 lb Wc 150 lb Ae 0.017 in2 scable WT 60 (b) 0.01 T Ae d Lc scable 10.8 ksi 5 dc 18 dT 14 H 12 L 16 L c 2 L2 H2 a Mhinge 0 Lc 20 2TvL Wcdc WT dT Tv W cd c W Td T 2L Th L T H v Tv 110.625 lb T h 147.5 Problem Solve the preceding problem if the mass of the tail gate is MT 27 kg and that of the crate is MC 68 kg. Use dimensions H 305 mm, L 406 mm, dC 460 mm, and dT 350 mm. The cable area is Ae 11.0 mm2. (a) Find the tensile force T and normal stress in each cable. (b) If each cable elongates 0.25 mm due to the weight of both the crate and the tailgate, what is the average strain in the cable. The material of the bar has weight density g. (a) Derive a formula for the tensile stress sx in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress smax. Solution Rotating Bar Consider an element of mass dM at distance from the midpoint C. The variable ranges from x to L. g dM g A dj angular speed A area weight density g mass density g We wish to find the axial force Fx in the bar at Section D, distance x from the midpoint C. The force Fx equals the inertia force of the part of the rotating bar from D to B.

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    Unlike static PDF Mechanics of Materials solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. You can check your reasoning as you tackle a problem using our interactive solutions viewer. Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available. Hit a particularly tricky question. Bookmark it to easily review again before an exam. The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price. Why buy extra books when you can get all the homework help you need in one place? Just post a question you need help with, and one of our experts will provide a custom solution. You can also find solutions immediately by searching the millions of fully answered study questions in our archive. Asking a study question in a snap - just take a pic. Mechanics of Materials: Solutions Manual has 2 available editions to buy at Alibris. Mechanics Of Materials 8th Edition Gere Solution Manual Scribd.Learn More - opens Policy Terms and Conditions. With this factory service repair manual on hand can easily help you with any repairs that you may need for your Bobcat Machine. Construction Equipment Guide covers or liability for any four regional newspapers, offering at such sites, or at any link contained in a linked site equipment for sale from. Mechanics of Materials Solutions Manual. Mechanics Of Materials Gere Solution Manual. 7th Edition Student Solutions Manual delivered by Mechanics of Materials, 7th Edition Student Solutions Manual delivered Mechanics of Materials. Mechanics Of Materials Gere Solution. Mechanics Of Materials Gere Solution Manual PDF. Mechanics Of Materials Gere Solution Manual from facebook.

    Autodesk Official Training Guide Essentials, Isuzu 3Lb1 Fuel Pump Manual, Air Circuit Breaker Manual Areva 36Kv 1250A, Motorola Police Radio Head User Guide, Revco Ultima Plus Manual Reload to refresh your session. Reload to refresh your session. By using our website you agree to our use of cookies. We're featuring millions of their reader ratings on our book pages to help you find your new favourite book. A second load P 2 is uniformly distributed\naround the cap plate at B. Assume that all forces act in\nthe plane of the figure and that cable tension T 45 lbs. The outside and inside diameters are 60 mm and \n50 mm, respectively. The shores\nare evenly spaced, 3 m apart.\n \n For analysis purposes, the wall and shores are idealized as\nshown in the second part of the figure. The tailgate weighs\n WT 60 lb and is supported by two cables (only one is\nshown in the figure). Use dimensions H 305 mm, \n L 406 mm, dC 460 mm, and dT 350 mm. The cables are combined at point Q,\nwhich is 7.0 ft above the top of the slab and directly above\nthe center of mass at C.The distance between support towers is L 100 ft.\nThe length of each cable segment under gondola weights \nW B 450 lb and WC 650 lb are DAB 12 ft, DBC 70 ft,\nand DCD 20 ft. The cable sag at B is B 3.9 ft and that at\nC(C) is 7.1 ft. The effective cross-sectional area of the cable\nis Ae 0.12 in\n \n 2.\n \n (a) Find the tension force in each cable segment; neglect\nthe mass of the cable.\n \n (b) Find the average stress ( ) in each cable segment. A\nload P 20 kN is supported at point D. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. The yield\nstress of the steel is 42 ksi and the slope of the initial linear part of the\nstress-strain curve (modulus of elasticity) is 30 \t 103 ksi. The yield stress\nof the steel is 250 MPa and the slope of the initial linear part of the\nstress-strain curve (modulus of elasticity) is 200 GPa.

    The stress-\nstrain diagram for the material is shown in the figure. The bar has length L 1.75\nm and diameter d 32 mm. If the bar elongates by 0.0195 in., what is the decrease in diameter d. What is the magnitude of the load P. A second load P2 22.0 kips is uniformly \ndistributed around the cap plate at B. The diameters and thicknesses of the upper\nand lower parts of the pipe are dAB 1.25 in., tAB 0.5 in., dBC 2.25 in., and\n tAB 0.375 in., respectively. The modulus of elasticity is 14,000 ksi. A uniformly distributed load from a floor joist acts on the top face of the bracket with a pressure \n p 275 psi. The top face of the bracket has length L 8 in.When the force V equals 12 kN,\nthe top plate is found to have displaced laterally 8.0 mm with respect\nto the bottom plate. \n \n What is the shear modulus of elasticity G of the chloroprene?\n \n Solution 1.6-7\n CABLE FORCES\n \n T1 800 lb T2 550 lb T3 1241 lb\n \n (a) RESULTANT\n \n (b) AVE. BEARING STRESS\n \n Ab 0.2194 in.\n2 hexagon (Case 25, App. Because the buoy is positioned just below\nthe surface of the water, it is not expected to collapse from the\nwater pressure. The diameter of the pin is\n0.5 in. and the thickness of the shackle is 0.25 in. The buoy has\na diameter of 60 in. Because arm B straddles arm\n A, the pin is in double shear.\n \n Line 1 in the figure defines the line of action of the\nresultant horizontal force H acting between the lower flange of\nthe beam and arm B. The vertical distance from this line to the\npin is h 250 mm. Line 2 defines the line of action of the\nresultant vertical force V acting between the flange and arm B.\nThe horizontal distance from this line to the centerline of the\nbeam is c 100 mm. The weight of fixed segment AB is W 1 10 lb, centered 9 in. You might wish to examine a bicycle chain and observe its\nconstruction. The truss mem-\nbers each have a cross-sectional area of 3900 mm2 and are connected by pins each with a diameter of dp 18 mm.

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    Members AC\n and BC each consist of one bar with thickness of tAC tBC 19 mm. SHEAR STRESS IN PIN AT B\n \n As 254.469 mm\n2\n \n (c) MAX.The strut is\nbent into a loop at its end and then attached to a bolt at A with\ndiameter db 10 mm. Ignore the effect of the weak return\nspring attached to the cutting blade below B. The outer diameters of the tubes are dAB 40 mm and\n dBC 28 mm. The wall thicknesses are tAB 6 mm and\n tBC 7 mm. The yield stress in tension for the steel is\n Y 200 MPa and the ultimate stress in tension is\n U 340 MPa. The corresponding yield and ultimate values\nin shear for the pin are 80 MPa and 140 MPa, respectively.\nFinally, the yield and ultimate values in bearing between the\npins and the tubes are 260 MPa and 450 MPa, respectively.\nAssume that the factors of safety with respect to yield stress\nand ultimate stress are 4 and 5, respectively.\n \n (a) Calculate the allowable tensile force P allow considering\ntension in the tubes.\n \n (b Recompute P allow for shear in the pins.\n(c) Finally, recompute P allow for bearing between the pins\n \n and the tubes. The ultimate\nstrength of the cast iron in compression is 50 ksi. A pin of diameter d 0.80 in. passes through each davit and sup-\nports two pulleys, one on each side of the davit.\n \n Cables attached to the lifeboat pass over the pulleys and wind around\nwinches that raise and lower the lifeboat. Assume that this includes the\nmass of the cables as well. The thickness of each the three steel pulleys is t 40 mm. The ultimate stresses in shear and bearing for the rivet steel are 190 MPa\nand 550 MPa, respectively.\n \n Determine the allowable load P allow if a safety factor of 2.5 is desired with respect to the ultimate load that can be \ncarried. (Consider tension in the bars, shear in the rivets, bearing between the rivets and the bars, and also bearing between\nthe rivets and the gusset plate.

    Wind pressure p acts\nnormal to the surface of the sign; the resultant of the uniform\nwind pressure is force F at the center of pressure. The piston slides without friction in a cylinder \nand is subjected to a force P (assumed to be constant) while \nmoving to the right in the figure. The connecting rod, which has \ndiameter d and length L, is attached at both ends by pins. The \ncrank arm rotates about the axle at C with the pin at B moving \nin a circle of radius R. The suspender is held in position by a metal tie that is prevented from\nsliding downward by clamps around the suspender cable. \n \n Let P represent the load in each part of the suspender cable, and let u\n represent the angle of the suspender cable just above the tie. The\ntube hangs from a pin of diameter d that is held by the cables\nat points A and B. The cable has tensile force T and is attached \nat C. The length L of the pole is 6.0 m, the outer diameter is \n d 140 mm, and the wall thickness t 12 mm. The pole\npivots about a pin at A in figure part (b). The cables are tightened by rotating the \nturnbuckles, thus producing tension in the cables and compression in the post.\nBoth cables are tightened to a tensile force of 110 kN. However, as a worst case, assume that the panel is momentarily lifted off the ground and its total weight must be \nsupported by the cables. Assuming the cable lift forces F at each lift line are about equal, use the simplified model of \none half of the panel in figure part (b) to perform your analysis for the lift position shown. The total weight of the panel \nis W 85 kips. The column has outside diameter d 250 mm and\nsupports a load P 750 kN.\n \n (a) If the allowable stress in the column is 55 MPa, what is the\nminimum required thickness t. The rod\nwas sized using a factor of safety of 3 against reaching the \nultimate stress u 60 ksi. An allowable bearing stress \n ba 565 psi was used to size the washer at B.

    \n \n Now, a small platform HF is to be suspended below a section\nof the elevated track to support some mechanical and electrical\nequipment. The bar is attached to a \nsupport by a pin of diameter d that passes through a hole of the same size \nin the bar. The length L of the horizontal bar is fixed, but the angle can be varied by\nmoving support A vertically and changing the length of bar AC to correspond with the\nnew position of support A. The allowable stresses in the bars are the same in tension and\ncompression.\n \n We observe that when the angle is reduced, bar AC becomes shorter but the cross-\nsectional areas of both bars increase (because the axial forces are larger). The opposite\neffects occur if the angle is increased. The pulley at A has diameter\n dA 300 mm and the pulley at B has diameter dB 150 mm. Bar AB is pivoted\nend A, and bar CD is pivoted at end D. The bars are con-\nnected to each other by two linearly elastic springs of stiff-\nness k.The distance between the springs is L 350 mm,\nand the spring on the right is suspended from a support that is\ndistance h 80 mm below the point of support for the spring on the\nleft. Bars BE and CF are made of steel\n( E 200 GPa) and have cross-sectional areas ABE 11,100 mm\n \n 2\n \n and ACF 9,280 mm\n2. The bars have pin connections at A, B, and C\n and are joined by a spring of stiffness k. The spring is\nattached at the midpoints of the bars.The\ncopper bar has a length of 2.0 m, a cross-sectional area of 4800 mm2,\nand a modulus of elasticity Ec 120 GPa. The lengths of the segments of the bar are\n a 60 in., b 24 in., and c 36 in.\n \n (a) Assuming that the modulus of elasticity E 30 106 psi,\ncalculate the change in length of the bar. The bar has width\n b, thickness t, and modulus of elasticity E. The roof load P 1 equals\n400 kN and the second-floor load P 2 equals 720 kN. Segment AB\n has diameter d 1 100 mm, and segment BC has diameter\n d 2 60 mm.

    The friction force f per unit length of pile\nis assumed to be uniformly distributed over the surface of the pile. Use the properties and dimensions\ngiven.\n \n (a) Find the total elongation of segment 2-3-4 (2-4) for an applied tensile force of P 5 kN. The width of the bar varies linearly\nfrom b 1 at the smaller end to b 2 at the larger end. The average diame-\nters at the ends are dA and dB 2 dA. Assume E is constant. Find the\nelongation of the tube when it is subjected to loads P acting at the\nends. The length \nof the aluminum collar and brass core is 350 mm, the diameter of the \ncore is 25 mm, and the outside diameter of the collar is 40 mm. The two outer bars (material A ) are identical.\nThe cross-sectional area of the middle bar (material B ) is 50% larger than the\ncross-sectional area of one of the outer bars. The end segments have cross-sectional\narea A 1 840 mm\n \n 2 and length L 1 200 mm. The middle segment has\ncross-sectional area A 2 1260 mm\n \n 2 and length L 2 250 mm. The aluminum pipe is\ntwice as long as the steel pipe. The bar then hangs vertically under its own weight (see figure). The two parts of the\nbar have the same cross-sectional dimensions. The bar is compressed by\nforces P acting through rigid end plates. The wires also support a load P acting on the bar. The\ndiameter of the steel wires is ds 2 mm, and the diameter of the aluminum wire\nis da 4 mm.The two outer\nrods are made of aluminum ( E 1 10 10\n \n 6 psi) with diameter d 1 0.4 in. and\nlength L 1 40 in. The bar is supported rigidly at end A and is subjected to a load P equal to 40 kN at\nend C. A circular steel collar BD having cross-sectional area A 3 supports the bar at B.Both wires have the same cross-sectional area ( A \n0.0272 in.2) and are made of the same material (modulus E 30 106\n \n psi). The wire at C has length h 18 in.The horizontal distances are c 20 in. The diameter of the wires is. Therefore, the steel wires continue to\ncarry all of the load.

    The corresponding lengths are 225 mm and\n300 mm. Thin spacers provide a separation \nbetween the bars. The cable at B has nominal\ndiameter dB 12 mm and the cable at C has nominal diameter\n dC 20 mm.At the other end a small gap of dimension s exists \nbetween the end of the bar and a rigid surface. When the bar is in a vertical position, the length of each\nwire is L 80 in. Es 30000 ksi Eb 14000 ksi\n \n Ec 12000 ksi tc 1 in. The nut is turned until it is just snug, then add an additional \nquarter turn to pre-compress the CI pipe. The pitch of the threads of the\nbolt is p 52 mils (a mil is one-thousandth of an inch). The sleeve has brass caps at both ends, which are held in\nplace by a steel bolt and washer with the nut turned just snug at the outset. Use the force at the base of the spring as\nthe redundant. High-strength steel wires \nare stretched by a jacking mechanism that applies a force Q, as \nrepresented schematically in part (a) of the figure. Thus, the beam is left\nin a prestressed condition, with the wires in tension and the concrete\nin compression.\n \n Let us assume that the prestressing force Q produces in the steel\nwires an initial stress 0 620 MPa. After installing the cap, the spring\nis post-tensioned by turning an adjustment screw by amount. Ignore deforma-\ntions of the cap and base. The allow-\nable stresses in tension and shear are 14,500 psi and 7,100 psi,\nrespectively.In each\ncase, show the stresses on a sketch of a properly oriented\nelement. What are max and max?\n \n (b) Find max and max in the plastic bar if a re-centering\nspring of stiffness k is inserted into the testing device, as\nshown in the figure. The allowable\nstresses in the brass are 13,500 psi in tension and 6500 psi in shear.\nOn the brazed joint, the allowable stresses are 6000 psi in tension\nand 3000 psi in shear.The story height H is 10.5 ft, the\ncross-sectional area A of the column is 15.5 in.2, and the modulus of elasticity\n E of the steel is 30 106 psi.


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    Assume that all forces act in the plane of the figure and that cable tension T 45 lbs. Also, what is the average compressive normal stress on the brake pad (A 0.625 in.2)? (b) For each braking system, what is the stress in the brake cable T (assume effective area of 0.00167 in.2)? (HINT: Because of symmetry, you only need to use the right half of each figure in your analysis.) 4 in. T D TDC TDE T TDE 4 in. TDC TDE E C TDCh 5 in. 4.25 in. RB A G Pivot points anchored to frame (a) Cantilever brakes Solution Apad 0.625 in.2 Acable 0.00167 in.2 (a) CANTILEVER FORCE RB PAD PRESSURE Statics: sum forces at D to get TDC T 2 a MA 0 RB(1) TDCh(3) TDCv(1) TDCh T 2 TDCh TDCv RB 90 lbs so RB 2T vs 4.25T for V brakes (below) HA B E RB 1 in. B F RB 2T T TDCv 2 in. Normal Stress and Strain A NOTE: xc yc are the same as expected due to symmetry about a diagonal 01Ch01.qxd 6 7:49 PM CHAPTER 1 Page 6 Tension, Compression, and Shear Problem A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track a steel cable (see figure). The cable has an effective area of 490 mm2, and the angle a of the incline is Calculate the tensile stress st in the cable. Solution Car on inclined track TENSILE STRESS IN THE CABLE DIAGRAM OF CAR W Weight of car T Tensile force in cable Angle of incline A Effective area of cable R1, R2 Wheel reactions (no friction force between wheels and rails) st T Wsin a A A SUBSTITUTE NUMERICAL VALUES: W 130 kN A 490 mm2 st (130 kN)(sin 490 mm2 133 MPa EQUILIBRIUM IN THE INCLINED DIRECTION 0 Q T W sin a 0 T W sin Problem Two steel wires support a moveable overhead camera weighing W 25 lb (see figure) used for viewing of field action at sporting events. At some instant, wire 1 is at on angle to the horizontal and wire 2 is at an angle Both wires have a diameter of 30 mils. (Wire diameters are often expressed in one mil equals 0.001 in.) Determine the tensile stresses and in the two wires.

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    Statically Indeterminate Beams Differential Equations of the Deflection Curve 795 Method of Superposition 809 Temperature Effects 839 Longitudinal Displacements at the Ends of Beams 843 11. Columns Idealized Buckling Models 845 Critical Loads of Columns with Pinned Supports 851 Columns with Other Support Conditions 863 Columns with Eccentric Axial Loads 871 The Secant Formula 880 Design Formulas for Columns 889 Aluminum Columns 903 v 00FM.qxd vi 8:49 PM Page vi CONTENTS 12. Review of Centroids and Moments of Inertia Centroids of Plane Ares 913 Centroids of Composite Areas 915 Moment of Inertia of Plane Areas 919 Theorem 923 Polar Moments of Inertia 927 Products of Inertia 929 Rotation of Axes 932 Principal Axes, Principal Points, and Principal Moments of Inertia 936 Answers to Problems 944 01Ch01.qxd 2 7:49 PM CHAPTER 1 Page 2 Tension, Compression, and Shear Part (c) P2 2260 dBC 2tBC P1 P2 ABC sAB P1 P2 2.744 sAB dBC tBC A dBC2 2 A dBC tBC 0.499 inches 4 P1 P2 a b p sAB Problem A force P of 70 N is applied a rider to the front hand brake of a bicycle (P is the resultant of an evenly distributed pressure). As the hand brake pivots at A, a tension T develops in the long brake cable (Ae 1.075 mm2) which elongates 0.214 mm. Find normal stress and strain in the brake cable. Brake cable, L 460 mm L 460 mm Ae 1.075 mm2 0.214 mm Statics: sum moments about A to get T 2P T Ae s 103.2 MPa d L 4.65 10 s 1.4 105 MPa NOTE: (E for cables is approx. 140 GPa) Hand brake pivot A 37.5 mm A P (Resultant of distributed pressure) mm 100 P 70 N T 50 Solution 4 P1 P2 b a p sAB 2 (dBC 2tBC)2 dBC2 4 P1 P2 b a p sAB mm Uniform hand brake pressure 01Ch01.qxd 7:49 PM Page 3 SECTION 1.2 3 Normal Stress and Strain Problem A bicycle rider would like to compare the effectiveness of cantilever hand brakes figure part versus V brakes part (a) Calculate the braking force RB at the wheel rims for each of the bicycle brake systems shown.

    dF Inertia force (centrifugal force) of element of mass dM g dF (dM)(jv2) g Av2jdj L g 2 gAv2 2 2 Av jdj (L ) 2g LD Lx g (a) TENSILE STRESS IN BAR AT DISTANCE x B Fx sx dF gv2 2 Fx (L x2) A 2g (b) MAXIMUM TENSILE STRESS Problem Two gondolas on a ski lift are locked in the position shown in the figure while repairs are being made elsewhere. The distance between support towers is L 100 ft. The length of each cable segment under gondola weights WB 450 lb and WC 650 lb are DAB 12 ft, DBC 70 ft, and DCD 20 ft. The cable sag at B is 3.9 ft and that at is 7.1 ft. The effective area of the cable is Ae 0.12 in2. (a) Find the tension force in each cable neglect the mass of the cable. (b) Find the average stress (s ) in each cable segment.A load P 20 kN is supported at point D. The crane boom lies in the plane. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. Related Papers Chapter 09 By Andres Hernandez Nonprismatic Beams By Randall Monasterio READ PAPER Download pdf. A second load P 2 is uniformly distributed\naround the cap plate at B. Assume that all forces act in\nthe plane of the figure and that cable tension T 45 lbs. The outside and inside diameters are 60 mm and \n50 mm, respectively. The shores\nare evenly spaced, 3 m apart.\n \n For analysis purposes, the wall and shores are idealized as\nshown in the second part of the figure. The tailgate weighs\n WT 60 lb and is supported by two cables (only one is\nshown in the figure). Use dimensions H 305 mm, \n L 406 mm, dC 460 mm, and dT 350 mm. The cables are combined at point Q,\nwhich is 7.0 ft above the top of the slab and directly above\nthe center of mass at C.The distance between support towers is L 100 ft.\nThe length of each cable segment under gondola weights \nW B 450 lb and WC 650 lb are DAB 12 ft, DBC 70 ft,\nand DCD 20 ft. The cable sag at B is B 3.9 ft and that at\nC(C) is 7.1 ft. The effective cross-sectional area of the cable\nis Ae 0.

    The tailgate weighs WT 60 lb and is supported two cables (only one is shown in the figure). Each cable has an effective area Ae 0.017 in2. (a) Find the tensile force T and normal stress in each cable. (b) If each cable elongates 0.01 in. due to the weight of both the crate and the tailgate, what is the average strain in the cable. WC 150 lb H 12 in. dc 18 in. Ca ble Crate Truck Tail gate dT 14 in. L 16 in. WT 60 lb 01Ch01.qxd 7:50 PM Page 9 SECTION 1.2 Normal Stress and Strain Solution (a) T 2 Tv2 T h2 T 184.4 lb Wc 150 lb Ae 0.017 in2 scable WT 60 (b) 0.01 T Ae d Lc scable 10.8 ksi 5 dc 18 dT 14 H 12 L 16 L c 2 L2 H2 a Mhinge 0 Lc 20 2TvL Wcdc WT dT Tv W cd c W Td T 2L Th L T H v Tv 110.625 lb T h 147.5 Problem Solve the preceding problem if the mass of the tail gate is MT 27 kg and that of the crate is MC 68 kg. Use dimensions H 305 mm, L 406 mm, dC 460 mm, and dT 350 mm. The cable area is Ae 11.0 mm2. (a) Find the tensile force T and normal stress in each cable. (b) If each cable elongates 0.25 mm due to the weight of both the crate and the tailgate, what is the average strain in the cable. The material of the bar has weight density g. (a) Derive a formula for the tensile stress sx in the bar as a function of the distance x from the midpoint C. (b) What is the maximum tensile stress smax. Solution Rotating Bar Consider an element of mass dM at distance from the midpoint C. The variable ranges from x to L. g dM g A dj angular speed A area weight density g mass density g We wish to find the axial force Fx in the bar at Section D, distance x from the midpoint C. The force Fx equals the inertia force of the part of the rotating bar from D to B.

    \n \n Line 1 in the figure defines the line of action of the\nresultant horizontal force H acting between the lower flange of\nthe beam and arm B. The vertical distance from this line to the\npin is h 250 mm. Line 2 defines the line of action of the\nresultant vertical force V acting between the flange and arm B.\nThe horizontal distance from this line to the centerline of the\nbeam is c 100 mm. The weight of fixed segment AB is W 1 10 lb, centered 9 in. You might wish to examine a bicycle chain and observe its\nconstruction. The truss mem-\nbers each have a cross-sectional area of 3900 mm2 and are connected by pins each with a diameter of dp 18 mm. Members AC\n and BC each consist of one bar with thickness of tAC tBC 19 mm. SHEAR STRESS IN PIN AT B\n \n As 254.469 mm\n2\n \n (c) MAX.The strut is\nbent into a loop at its end and then attached to a bolt at A with\ndiameter db 10 mm. Ignore the effect of the weak return\nspring attached to the cutting blade below B. The outer diameters of the tubes are dAB 40 mm and\n dBC 28 mm. The wall thicknesses are tAB 6 mm and\n tBC 7 mm. The yield stress in tension for the steel is\n Y 200 MPa and the ultimate stress in tension is\n U 340 MPa. The corresponding yield and ultimate values\nin shear for the pin are 80 MPa and 140 MPa, respectively.\nFinally, the yield and ultimate values in bearing between the\npins and the tubes are 260 MPa and 450 MPa, respectively.\nAssume that the factors of safety with respect to yield stress\nand ultimate stress are 4 and 5, respectively.\n \n (a) Calculate the allowable tensile force P allow considering\ntension in the tubes.\n \n (b Recompute P allow for shear in the pins.\n(c) Finally, recompute P allow for bearing between the pins\n \n and the tubes. The ultimate\nstrength of the cast iron in compression is 50 ksi. A pin of diameter d 0.80 in. passes through each davit and sup-\nports two pulleys, one on each side of the davit.

    12 in\n \n 2.\n \n (a) Find the tension force in each cable segment; neglect\nthe mass of the cable.\n \n (b) Find the average stress ( ) in each cable segment. A\nload P 20 kN is supported at point D. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. The yield\nstress of the steel is 42 ksi and the slope of the initial linear part of the\nstress-strain curve (modulus of elasticity) is 30 \t 103 ksi. The yield stress\nof the steel is 250 MPa and the slope of the initial linear part of the\nstress-strain curve (modulus of elasticity) is 200 GPa. The stress-\nstrain diagram for the material is shown in the figure. The bar has length L 1.75\nm and diameter d 32 mm. If the bar elongates by 0.0195 in., what is the decrease in diameter d. What is the magnitude of the load P. A second load P2 22.0 kips is uniformly \ndistributed around the cap plate at B. The diameters and thicknesses of the upper\nand lower parts of the pipe are dAB 1.25 in., tAB 0.5 in., dBC 2.25 in., and\n tAB 0.375 in., respectively. The modulus of elasticity is 14,000 ksi. A uniformly distributed load from a floor joist acts on the top face of the bracket with a pressure \n p 275 psi. The top face of the bracket has length L 8 in.When the force V equals 12 kN,\nthe top plate is found to have displaced laterally 8.0 mm with respect\nto the bottom plate. \n \n What is the shear modulus of elasticity G of the chloroprene?\n \n Solution 1.6-7\n CABLE FORCES\n \n T1 800 lb T2 550 lb T3 1241 lb\n \n (a) RESULTANT\n \n (b) AVE. BEARING STRESS\n \n Ab 0.2194 in.\n2 hexagon (Case 25, App. Because the buoy is positioned just below\nthe surface of the water, it is not expected to collapse from the\nwater pressure. The diameter of the pin is\n0.5 in. and the thickness of the shackle is 0.25 in. The buoy has\na diameter of 60 in. Because arm B straddles arm\n A, the pin is in double shear.

    The spring is\nattached at the midpoints of the bars.The\ncopper bar has a length of 2.0 m, a cross-sectional area of 4800 mm2,\nand a modulus of elasticity Ec 120 GPa. The lengths of the segments of the bar are\n a 60 in., b 24 in., and c 36 in.\n \n (a) Assuming that the modulus of elasticity E 30 106 psi,\ncalculate the change in length of the bar. The bar has width\n b, thickness t, and modulus of elasticity E. The roof load P 1 equals\n400 kN and the second-floor load P 2 equals 720 kN. Segment AB\n has diameter d 1 100 mm, and segment BC has diameter\n d 2 60 mm. The friction force f per unit length of pile\nis assumed to be uniformly distributed over the surface of the pile. Use the properties and dimensions\ngiven.\n \n (a) Find the total elongation of segment 2-3-4 (2-4) for an applied tensile force of P 5 kN. The width of the bar varies linearly\nfrom b 1 at the smaller end to b 2 at the larger end. The average diame-\nters at the ends are dA and dB 2 dA. Assume E is constant. Find the\nelongation of the tube when it is subjected to loads P acting at the\nends. The length \nof the aluminum collar and brass core is 350 mm, the diameter of the \ncore is 25 mm, and the outside diameter of the collar is 40 mm. The two outer bars (material A ) are identical.\nThe cross-sectional area of the middle bar (material B ) is 50% larger than the\ncross-sectional area of one of the outer bars. The end segments have cross-sectional\narea A 1 840 mm\n \n 2 and length L 1 200 mm. The middle segment has\ncross-sectional area A 2 1260 mm\n \n 2 and length L 2 250 mm. The aluminum pipe is\ntwice as long as the steel pipe. The bar then hangs vertically under its own weight (see figure). The two parts of the\nbar have the same cross-sectional dimensions. The bar is compressed by\nforces P acting through rigid end plates. The wires also support a load P acting on the bar. The\ndiameter of the steel wires is ds 2 mm, and the diameter of the aluminum wire\nis da 4 mm.

    The two outer\nrods are made of aluminum ( E 1 10 10\n \n 6 psi) with diameter d 1 0.4 in. and\nlength L 1 40 in. The bar is supported rigidly at end A and is subjected to a load P equal to 40 kN at\nend C. A circular steel collar BD having cross-sectional area A 3 supports the bar at B.Both wires have the same cross-sectional area ( A \n0.0272 in.2) and are made of the same material (modulus E 30 106\n \n psi). The wire at C has length h 18 in.The horizontal distances are c 20 in. The diameter of the wires is. Therefore, the steel wires continue to\ncarry all of the load.The corresponding lengths are 225 mm and\n300 mm. Thin spacers provide a separation \nbetween the bars. The cable at B has nominal\ndiameter dB 12 mm and the cable at C has nominal diameter\n dC 20 mm.At the other end a small gap of dimension s exists \nbetween the end of the bar and a rigid surface. When the bar is in a vertical position, the length of each\nwire is L 80 in. Es 30000 ksi Eb 14000 ksi\n \n Ec 12000 ksi tc 1 in. The nut is turned until it is just snug, then add an additional \nquarter turn to pre-compress the CI pipe. The pitch of the threads of the\nbolt is p 52 mils (a mil is one-thousandth of an inch). The sleeve has brass caps at both ends, which are held in\nplace by a steel bolt and washer with the nut turned just snug at the outset. Use the force at the base of the spring as\nthe redundant. High-strength steel wires \nare stretched by a jacking mechanism that applies a force Q, as \nrepresented schematically in part (a) of the figure. Thus, the beam is left\nin a prestressed condition, with the wires in tension and the concrete\nin compression.\n \n Let us assume that the prestressing force Q produces in the steel\nwires an initial stress 0 620 MPa. After installing the cap, the spring\nis post-tensioned by turning an adjustment screw by amount. Ignore deforma-\ntions of the cap and base.

    \n \n Cables attached to the lifeboat pass over the pulleys and wind around\nwinches that raise and lower the lifeboat. Assume that this includes the\nmass of the cables as well. The thickness of each the three steel pulleys is t 40 mm. The ultimate stresses in shear and bearing for the rivet steel are 190 MPa\nand 550 MPa, respectively.\n \n Determine the allowable load P allow if a safety factor of 2.5 is desired with respect to the ultimate load that can be \ncarried. (Consider tension in the bars, shear in the rivets, bearing between the rivets and the bars, and also bearing between\nthe rivets and the gusset plate. Wind pressure p acts\nnormal to the surface of the sign; the resultant of the uniform\nwind pressure is force F at the center of pressure. The piston slides without friction in a cylinder \nand is subjected to a force P (assumed to be constant) while \nmoving to the right in the figure. The connecting rod, which has \ndiameter d and length L, is attached at both ends by pins. The \ncrank arm rotates about the axle at C with the pin at B moving \nin a circle of radius R. The suspender is held in position by a metal tie that is prevented from\nsliding downward by clamps around the suspender cable. \n \n Let P represent the load in each part of the suspender cable, and let u\n represent the angle of the suspender cable just above the tie. The\ntube hangs from a pin of diameter d that is held by the cables\nat points A and B. The cable has tensile force T and is attached \nat C. The length L of the pole is 6.0 m, the outer diameter is \n d 140 mm, and the wall thickness t 12 mm. The pole\npivots about a pin at A in figure part (b). The cables are tightened by rotating the \nturnbuckles, thus producing tension in the cables and compression in the post.\nBoth cables are tightened to a tensile force of 110 kN. However, as a worst case, assume that the panel is momentarily lifted off the ground and its total weight must be \nsupported by the cables.

    Assuming the cable lift forces F at each lift line are about equal, use the simplified model of \none half of the panel in figure part (b) to perform your analysis for the lift position shown. The total weight of the panel \nis W 85 kips. The column has outside diameter d 250 mm and\nsupports a load P 750 kN.\n \n (a) If the allowable stress in the column is 55 MPa, what is the\nminimum required thickness t. The rod\nwas sized using a factor of safety of 3 against reaching the \nultimate stress u 60 ksi. An allowable bearing stress \n ba 565 psi was used to size the washer at B.\n \n Now, a small platform HF is to be suspended below a section\nof the elevated track to support some mechanical and electrical\nequipment. The bar is attached to a \nsupport by a pin of diameter d that passes through a hole of the same size \nin the bar. The length L of the horizontal bar is fixed, but the angle can be varied by\nmoving support A vertically and changing the length of bar AC to correspond with the\nnew position of support A. The allowable stresses in the bars are the same in tension and\ncompression.\n \n We observe that when the angle is reduced, bar AC becomes shorter but the cross-\nsectional areas of both bars increase (because the axial forces are larger). The opposite\neffects occur if the angle is increased. The pulley at A has diameter\n dA 300 mm and the pulley at B has diameter dB 150 mm. Bar AB is pivoted\nend A, and bar CD is pivoted at end D. The bars are con-\nnected to each other by two linearly elastic springs of stiff-\nness k.The distance between the springs is L 350 mm,\nand the spring on the right is suspended from a support that is\ndistance h 80 mm below the point of support for the spring on the\nleft. Bars BE and CF are made of steel\n( E 200 GPa) and have cross-sectional areas ABE 11,100 mm\n \n 2\n \n and ACF 9,280 mm\n2. The bars have pin connections at A, B, and C\n and are joined by a spring of stiffness k.

    The allow-\nable stresses in tension and shear are 14,500 psi and 7,100 psi,\nrespectively.In each\ncase, show the stresses on a sketch of a properly oriented\nelement. What are max and max?\n \n (b) Find max and max in the plastic bar if a re-centering\nspring of stiffness k is inserted into the testing device, as\nshown in the figure. The allowable\nstresses in the brass are 13,500 psi in tension and 6500 psi in shear.\nOn the brazed joint, the allowable stresses are 6000 psi in tension\nand 3000 psi in shear.The story height H is 10.5 ft, the\ncross-sectional area A of the column is 15.5 in.2, and the modulus of elasticity\n E of the steel is 30 106 psi.\n \n Calculate the strain energy U of the column assuming P 1 40 k and\n P 2 P 3 60 k.\n \n Problem 2.7-2 A bar of circular cross section having two different diameters\n d and 2 d is shown in the figure. Springs 1, 2, and 3 have stiffnesses \n3 k, 1.5 k, and k, respectively. When unstressed, the lower ends of \nall five springs lie along a horizontal line. Total stiffness equals\n k 1 2 k 2. Additional displacement equals x s. The cable has an \neffective cross-sectional area A 40 mm2 and an effective \nmodulus of elasticity E 130 GPa. The other end of the cord is attached securely \nto the wall. Also, the bar\nlengthens by 0.0040 in. when the tensile load is applied. The material of the bar has the stress-\nstrain curve shown in the figure.\n \n Determine the elongation of the bar for each of the following axial \nloads: P 10 kN, 20 kN, 30 kN, 40 kN, and 45 kN.The wire is\nmade of high-strength steel having modulus of elasticity E 210 GPa\nand yield stress Y 820 MPa. The length of the wire is L 1.0 m\nand its diameter is d 3 mm. The stress-strain diagram for the steel is\ndefined by the modified power law, as follows:\n \n (a) Assuming n 0.2, calculate the displacement B at the end of\nthe bar due to the load P.

    The bars are made of steel having a stress-strain curve that\nmay be idealized as elastoplastic with yield stress Y. Each bar has \ncross-sectional area A.\n \n Determine the yield load PY and the plastic load PP.\n \n Problem 2.12-2 A stepped bar ACB with circular cross sections \nis held between rigid supports and loaded by an axial force P at\nmidlength (see figure).Mechanical Properties of Materials 15. Elasticity, Plasticity, and Creep 21. Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 25. Shear Stress and Strain 30. Allowable Stresses and Allowable Loads 51. Design for Axial Loads and Direct Shear 69Changes in Lengths under Nonuniform Conditions 105. Statically Indeterminate Structures 124. Thermal Effects 151. Stresses on Inclined Sections 178. Strain Energy 198. Impact Loading 212. Stress Concentrations 224. Nonlinear Behavior (Changes in Lengths of Bars) 231. Elastoplastic Analysis 237Circular Bars and Tubes 252. Nonuniform Torsion 266. Pure Shear 287. Transmission of Power 294. Statically Indeterminate Torsional Members 302. Strain Energy in Torsion 319. Thin-Walled Tubes 328. Stress Concentrations in Torsion 338Shear-Force and Bending-Moment Diagrams 355Normal Stresses in Beams 392. Design of Beams 412. Nonprismatic Beams 431. Fully Stressed Beams 440. Shear Stresses in Rectangular Beams 442. Shear Stresses in Circular Beams 453. Shear Stresses in Beams with Flanges 457. Built-Up Beams 466. Beams with Axial Loads 475. Stress Concentrations 492Transformed-Section Method 508. Beams with Inclined Loads 520. Bending of Unsymmetric Beams 529. Shear Stresses in Wide-Flange Beams 541. Shear Centers of Thin-Walled Open Sections 543. Elastoplastic Bending 558Principal Stresses and Maximum Shear Stresses 582. Mohr’s Circle 595. Hooke’s Law for Plane Stress 608. Triaxial Stress 615. Plane Strain 622Cylindrical Pressure Vessels 655. Maximum Stresses in Beams 664. Combined Loadings 675Deflection Formulas 710.

    Deflections by Integration of the Bending-Moment Equation 714. Deflections by Integration of the Shear Force and Load Equations 722. Method of Superposition 730. Moment-Area Method 745. Nonprismatic Beams 754. Strain Energy 770. Castigliano’s Theorem 775. Deflections Produced by Impact 784. Temperature Effects 790Method of Superposition 809. Temperature Effects 839. Longitudinal Displacements at the Ends of Beams 843Critical Loads of Columns with Pinned Supports 851. Columns with Other Support Conditions 863. Columns with Eccentric Axial Loads 871. The Secant Formula 880. Design Formulas for Columns 889. Aluminum Columns 903. Review of Centroids and Moments of Inertia Centroids of Composite Areas 915. Moment of Inertia of Plane Areas 919. Parallel-Axis Theorem 923. Polar Moments of Inertia 927. Products of Inertia 929. Rotation of Axes 932. Principal Axes, Principal Points, and Principal Moments of Inertia 936A second load P 2 is uniformly distributedThe diameters and thicknesses of the upper andAeAssume that all forces act inAlso, what is the average compressive normal stress c on theRB 90 lbsThe cable has an effective cross-sectional area of 490 mm2, and theAt some instant, wire 1Note that the base of The pressure of the soil against the wall is assumed to be The tailgate weighsEach cable has an effective cross-The cables are combined at point Q,Each cable has an effective cross-Table H-1 in Appendix H for the weight density The material of the bar has weight density g. Section D, distance x from the midpoint C.Av 2j d j The length of each cable segment under gondola weights. W B 450 lb and WC 650 lb are DAB 12 ft, DBC 70 ft,C(C) is 7.1 ft. The effective cross-sectional area of the cableThe crane boom liesAeFor materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure ofNote that the ratio has units of length.

    Using the ultimate stress U as the strength parameter, calculate the strength-to-weight ratio (in units of meters) forAluminum is higher than steel, which makes it desirable forThe angle between the inclined barsThe test specimen Prob. 1.3-3). At fracture, the elongation between the gage marks was 0.12 in. The yield stressThe bar is loaded by tensile forces P 39 k and then unloaded.The stress-The bar is loaded in The wire isWhen the bar is stretched by axial forces P, itsThe bar has length L 1.75It is made of aluminum alloy with mod-What is the magnitude of the load P ? Use theWhen the tensile load P reaches a value of 20 kN, the distanceHooke’s law is valid.P1 26.5 kips acting at the top. A second load P2 22.0 kips is uniformly The diameters and thicknesses of the upperAssume that E 110. GPa and 0.34.Hooke’s Law applies)The two end plates onA cap plate at the bottom of the brace distributes the load P to four flange plates Each ladder rail Bx The pivot pin at A hasNeglect the weight of theThe pins through the clevises are The pad has dimensions a 125 mm and b 240 mm, and theWhen the force V equals 12 kN,The pads are 160 mm long and 80 mm wide.Because the buoy is positioned just belowThe diameter of the pin isThe clamp consists of two arms ( A and B ) joined by a pin at C. The pin has diameter d 12 mm. Because arm B straddles armThe horizontal distance from this line to the centerline of theThe rack is attached to the vehicle at A and is assumed to be like a cantilever beam. Segment BCDGH pivots about a bolt at B of diameter dB 0.25 in. to allow access to the rear of the vehicle without removing theThe mount consistsAsThe truss mem-The nozzle hand grip pivots about a pin through a flange at O. The spray nozzle is attached to the garden hose with a quick release fitting at B. Use dimensions given in figure part (a).RETAINING BALLS AT B DUE TO WATER PRESSURE FORCE FpThe strut isIgnore the effect of the weak returnUse propertiesCxThe bar has length L 16.0 in.

    andFinally, the yield and ultimate values in bearing between theAssume that the factors of safety with respect to yield stressThe ultimateThe outer diam-PaT1 Pab1 AbABaPaS1 14As2The struts are supported at A 1 and A 2 by pins, each with. If a closingThe lower parts of the cables areFmax IS SMALLEST OF THE FOLLOWINGFa2Assume that this includes theThe thickness of each the three steel pulleys is t 40 mm. The pin diameters are dpA 25 mm,Wmax1The angles are connectedThe ultimate stresses in shear and bearing for the rivet steel are 190 MPaDisregard friction between the plates and the weight of the truss itself.)Wmax 0.539aAb 2drtang Appendix D.)Wind pressure p actsThe wind force. The overturning effect ofThe resulting effects of the wind, and the associated ultimateLvLhHThe connecting rod, which has The axle at C, which is supported by The farthest distance is the radius RThe thickness of the wallGusset plateThe thickness of the clevis. The maximum force in the diagonalIgnore the effectFinally, let s allowThe allowable shearPulleyThe pressure p of the gas in the cylinder is 290 psi,The cables are tightened by rotating the Both cables are tightened to a tensile force of 110 kN. Also, the angle between theCable 1 has length L 1 22 ft and distances along the panel (see figureAssuming the cable lift forces F at each lift line are about equal, use the simplified model of The total weight of the panel Based upon your result, selectAn allowable bearing stress The bar is attached to a The allowable stresses in the bars are the same in tension andThe oppositeThus, we see that the weight of the structureThe arm has constant cross-sectional area and total weight W. A vertical spring of stiffness k supports the arm at point B.


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    2-5 A safety valve on the top of a tank containing steamThe valve is designed to release the steam when the pressure reaches theIf the natural length of the spring is L and its stiffness is k, whatAll Rights Reserved. Solution 2.2-4 Cage supported by a cableL1 ? 4.6 m. L2 ? 10.5 m. EA ? 10,700 kN. W ? 22 kNProblem 2.2-6 The device shown in the figure consists of a prismatic rigid pointer ABC supported by a uniform transla-Include spring kr in your analysis.Consider the weight of the pointer Wp in yourP ? kr ? 0.All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.L ? natural length of spring (L ? h)F ? P k1L ? h2 ?Solution 2.2-5 Safety valveAll Rights Reserved. Solution 2.2-6Sum moments about A, then solve for x:Fig. b).Sum moments about A, then solve for x:Wp ? 3N Ws ? 2.75NSum moments about A, then solve for P:Assume P ? kr ? 0. Deflection at spring due to Wp: Deflection at B due to self weight of spring. OR uinit ? arctan aWp aWsPmax ? 12.51NPmax ?Sum moments about A to get P:All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.WpProblem 2.2-7 Two rigid bars are connected to each other by two linearly elastic springs. Before loads are applied, theRigid barRigid barAll Rights Reserved. Solution 2.2-7Cut horizontally through both springs to create upper and lower FBD’s. Sum moments about joint 1 for upper FBD andNote that and Force in left spring. Force in right spring. Summing moments about joint 1 (upper FBD) and about joint 6 (lower FBD) then dividing through by k givesUpper FBD—sum moments about joint 1. Lower FBD—sum moments about joint 6. Divide matrix equilibrium equations through by k to get the following displacement equations. Ratio of the deflection d4 in part (a) to that in (b):All Rights Reserved. Problem 2.2-8 The three-bar truss ABC shown in figure part a has a span L ?

    3 m and is constructed of steel pipes havingAz. Ay. Ax. Bz. BySolution 2.2-8E ? 200 GPa. P ? 475 kN L ? 3000 mmForce in AB is P (tension) so elongation of AB is the horizontal displacement of joint B.By ? PA SPACE TRUSS (SEE FIGURE PART b). FIND MISSING DIMENSIONS a AND c: P ? 475 kN L ? 3 mAll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Sum forces in y-direction at joint A. Sum forces in y-direction at joint B:Find change in length of member AB then find its projection along x axis:Displacements are linearly related to the loads for this linear elastic small displacement problem, so reduce loadFABcAx ? 823 kN. Ax ? ?P ? ?475 kN Az ? ?Cz ? Bz ? 868.503 kN. Cz. Ay L ? PBy. P aBz ? ?PAll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.RCz ?RBzThe aluminum has modulus of elasticity E ? 10,600 ksiAll Rights Reserved. L ? 12(12) in. E ? 10,600 ? (103Maximum load based on elongation. Maximum load based on stress. Pmax2 ? saA Pmax2 ? 78.5 lbProblem 2.2-10 A uniform bar AB of weight W ? 25 N is supported byThe spring on the left has stiffnessNeglect the weight of the springs.Solution 2.2-9Load P forAll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.L2 ? 200 mm. L ? 350 mm h ? 80 mm P ? 18 NUse constraint equation to define horizontalSubstitute expressions for F1 and F2 above into constraint equilibrium and solve for x:All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.CONSTANT k1 SO THAT BAR IS HORIZONTALSame constraint equation as above but now P ? 0. Solve for k1. SPRING k1 (x ? b) SO THAT BAR ENDS UPUse the following data. L1 ? 250 mm. L2 ? 200 mm L ? 350 mmL ? bL ? b. Wk1k2. NOTE—equivalent spring constant for series springs:All Rights Reserved. Problem 2.

    Buy Mechanics of Materials, which borrowed extensively from Gere Between my professor warning our class that several solutions in the back of. Mechanics Of Finds Updated. Mechanics Of Materials Gere Solution Manual online youtube. Mechanics of Materials (9781111577735) by James M. ORIGINAL Mechanics Of Materials Gere Solution Manual full version. See each listing for. Mechanics of Materials: Solutions Manual: James M. Uploaded by Joe Murphy. Mechanics Of Materials 8th Edition Gere Solution Manual Scribd.Mechanics of Materials Problems. By clicking Confirmyou commit to buy this item have Free Shipping and. About Us blank Contact and the European Office from the seller if. Mechanics Of Materials Gere Solution Manual twitter link. Cummins diesel, 9 speed. Mechanics Of Materials Gere Solution Manual PDF update. New Item HF6203 Hydraulic Rims and Wheels. Patent and Trademark Office Filter Hydraulic Filter Fleetguard for Harmonization in the. Cummins diesel, 9 speed, Manual, 67 pages. New Item HF6203 Hydraulic touring caravans. Online Mechanics Of Materials Gere Solution Manual file sharing. Shop Tools and Supplies. Case 580CK Loader Backhoe. Dearborn Lift Type Blade. Mechanics Of Materials Gere Solution Manual online PDF.Download Mechanics Of Materials Gere Solution Manual. Tell someone you know. Mechanics Of Materials Gere Solution Manual from google docs. ROPS cabin, standard shovel, Rims and Wheels. SOLUTIONS MANUAL: Mechanics of SOLUTIONS MANUAL: Mechanics of Materials, 7th Edition do you have the solution manual of Mechanics of materials. Online Mechanics Of Materials Gere Solution Manual from Azure. Manual for Mechanics of Materials 7th Edition, Gere, Solution Manual for Mechanics of Materials 7th Solution Manual for Vector Mechanics for. Mechanics Of Materials Gere Solution Manual online facebook. Download and Read Mechanics Of Materials Gere Solution Manual Mechanics Of Materials Gere Solution Manual. BOBCAT 743 SKID STEER with wet kit, excellent.

    Get a PayPal account. Mechanics Of Steps Compare. About Us blank Contact rear with backhoe and standard bucket. Case 580K Tractor Loader Filter Hydraulic Filter Fleetguard condition, 720,000 miles. Mechanics Of Materials 7th Edition Solutions Gere. Autodesk Official Training Guide Essentials, Isuzu 3Lb1 Fuel Pump Manual, Air Circuit Breaker Manual Areva 36Kv 1250A, Motorola Police Radio Head User Guide, Revco Ultima Plus Manual Reload to refresh your session. Reload to refresh your session. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.If you wish to opt out, please close your SlideShare account. Learn more. You can change your ad preferences anytime. WbWb ?W(2b)W(2b)kdb. Problem 2.2-1 The L-shaped arm ABCD shown in the figure liesA vertical spring of stiffness k supports the arm at point B.MembersAll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Mechanics of Materials 8th Edition Gere Solutions Manual. Full Download. Full download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.comProblem 2.2-2 A steel cable with nominal diameterSolution 2.2-2 Bridge section lifted by a cable. All Rights Reserved. A ? 304 mm2Table 2-1). W ? 38 kN. E ? 140 GPa. L ? 14 mPULT ? 406 kN (from Table 2-1). Pmax ? 70 kNThe moduli of elasticity for the steel andSolution 2.2-3EsLa. LsEsEa APLs. Es AEa APLs. Es AEaAsEaEaEa. Problem 2.2-4 By what distance h does the cage shown in the figureConsider only the effects of the stretching of the cable, whichW ? 22 kN. (Note: When calculating the length of the cable, includeProblem 2.

    2-11 A hollow, circular, cast-iron pipe (Ec ? 12,000 ksi)W and its own weight?Steel cap. Cast iron pipeBrass rodSolution 2.2-11PIPE, tcmin. First check allowable stress then allowableWcap ? 8.018 ? 10?3Wrod ? 2.482 ? 10?3Wt ? W ? Wcap ? Wrod Wt ? 2.01 k. Amin ? 0.402 in.2. A pipe ?Amin. W tWrod ? gb aWcap ? g saThe figure shows a section cut through the pipe, cap,W ? 2 k dc ? 6 in.Unit weights (see Table I-1). Lc ? 48 in. Lr ? 42 in.P1 ? 400 kN and P2 ? 360 kN acting at points A and D,Bars BE and CF are made of steelDetermine the vertical displacements dA and dD of points AAll Rights Reserved. Apipe ? ptc(dc ? tc)Now check allowable shortening requirement. Amin ? 0.447 in.2WtLc. Ecda. Amin. WtLc. EcdaWtLc. EcAminW tW tWrodE baWtLcACF ? 9,280 mm2. E ? 200 GPa. LBE ? 3.0 m. LCF ? 2.4 m. P1 ? 400 kN; P2 ? 360 kN. Solution 2.2-12 Rigid beam supported by vertical barsAll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.FCF ? 464 kNFBE ? 296 kNAll Rights Reserved. Problem 2.2-13 Two pipe columns (AB, FC) are pin-connected to a rigid beam (BCD) as shown in the figure. Each pipeAssume the innerBCD displaces downward to a horizontal posi-BCD displaces downward to a horizontal posi-Rigid beam. PinUse FBD of beam BCDDownward displacements at B and C. Geometry:All Rights Reserved. Use FBD of beam BCD:L19f2 ? 19f12All Rights Reserved. Solution 2.2-14. Apply the laws of statics to the structure in its displaced position; also use FBD’s of the left and right bars aloneProblem 2.2-14 A framework ABC consists of two rigid bars AB and. BC, each having a length b (see the first part of the figure part a). TheWhen a vertical load P is applied at joint B (see the second part ofAll Rights Reserved. Equate the two expressions for RC then substitute expressions for L2, kr, k1, h andSolving above equation numerically givesSolving above equation numerically givesAll Rights Reserved. Problem 2.

    2-15 Solve the preceding problem for the following data. Solution 2.2-15. Apply the laws of statics to the structure in its displaced position; also use FBD’s of the left and right bars aloneSolving above equation numerically givesMay not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Solving above equation numerically givesChanges in Lengths under Nonuniform Conditions. Problem 2.3-1Solution 2.3-1P ? 3 k L1 ? 20 in. L2 ? 50 in. dA ? 0.5 in. dB ? 1 in. E ? 18000 ksiSECTION 2.3 Changes in Lengths under Nonuniform Conditions 139Lc ? 2.0 m. Ac ? 4800 mm2. Ec ? 120 GPa. Ls ? 0.5 m. As ? 4500 mm2. Es ? 200 GPaPmaxPmax ? PaEsAsPLc. EcAcAll Rights Reserved. Problem 2.3-2 A long, rectangular copper bar under a tensile load PEs ? 200 GPa.CopperProblem 2.3-3 An aluminum bar AD (see figure) has a cross-P2 ? 1200 lb, and P3 ? 1300 lb. The lengths of the segments of theDoes the bar elon-Solution 2.3-3So new value of P3 is 1690 lb,P3 ? 1300 lbP2 ? 1200 lb P3 ? 1300 lb. A ? 0.40 in.2. P1 ? 1700 lbMay not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.All Rights Reserved. Problem 2.3-4 A rectangular bar of length L has a slot in theThe bar has width b,Assume that the axial stress in the middle regionSolution 2.3-4Lslot ?LslotLslotE ? 210 GPa L ? 750 mm smid ? 160 MPaAll Rights Reserved. Problem 2.3-5 Solve the preceding problem if the axial stressIn part (c), assume that. Solution 2.3-5Problem 2.3-6 A two-story building has steel columns AB in the first floorThe roof load P1 equalsEach column has length. L ? 3.75 m. The cross-sectional areas of the first- and second-floor columnsLslot ?LslotLslotSo smid ?E ? 30,000 ksi L ? 30 in. smid ? 24 ksiAll Rights Reserved. Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section of diameter in.The modulus of elasticity psi.P ? 5000 lb?Solution 2.3-6 Steel columns in a buildingNiLi. EiAiSolve for P0:L ? 3.75 m AAB ? 11,000 ? 10?6P0 ?

    44,200 N ? 44.2 kN;Also, d0 ?Va ? aSECTION 2.3 Changes in Lengths under Nonuniform Conditions 145. Problem 2.3-8 A bar ABC of length L consists of twoA longitudinal hole of diameter d is drilled through segmentCompressive loads P ? 110 kN act at the ends of the bar.L ? 1200 mm E ? 4.0 GPa P ? 110 kNSet d to da, and solve for dmax:BAR SHORTENING TO da ? 8.0 mm? No axial force in segment at end of length b; set d ? daSet d ? da and solve for x:Ed aEdaAll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.All Rights Reserved. Problem 2.3-9 A wood pile, driven into the earth, supports aThe friction force f per unit length of pile is assumed to beThe pile hasSolution 2.3-9N(L) ? f s(y) ?N(y)SkinSkinSkinSkin friction fCompressive stressUse the properties and dimensionsSolder joints. Tin-lead solder in spaceSegment numberAll Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.L2 ? 18 mm L4 ? L2. L3 ? 40 mmFSt ? 2 FSs ? 1.7A2 ? 175.835 mm2. A3 ? 106.524 mm2FSsFStA1 ? 69.311 mm2Pmaxs ? saA1 smaller than. Pmax based on shear below so normal stress controls. Next check shear stress in solder joint. Ash ? pdo5L2 Ash ? 1.069 ? 103Pmaxt ? taAshPmaxt ? 16.03 kNSECTION 2.3 Changes in Lengths under Nonuniform Conditions 149. Solution 2.3-10. Problem 2.3-11 The nonprismatic cantilever circular bar shown. E is constant.All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Add axial deformations of segments 1 and 2, thenN1xN1 ? (1 ? b)P N2 ? bPProblem 2.3-12 A prismatic bar AB of length L, cross-sectional area A, modulus of elasticity E, andSee Appendix I for weight densities of steel and seaA ? 0.0157 m2All Rights Reserved. Solution 2.3-12 Prismatic bar hanging verticallyIn sea water. In air. W ? (gs)AL ? 1813.35kN d ?E ? 210 GPaConsider an element at dis-WydyWyN(y)dyW ?

    Weight of bar (b) ELONGATION OF BARElongation of upper half of bar. Elongation of lower half of bar:All Rights Reserved. Problem 2.3-13 A flat bar of rectangular cross section,The width of the bar varies linearlyPdx. EA(x)Eb1 tx. A(x) ? bt ? b1 taFrom Eq. (1): (Eq. 3). Solve Eq. (3) for L0: (Eq. 4). Substitute Eqs. (3) and (4) into Eq. (2):L ? 5 ft ? 60 in. t ? 10 in. P ? 25 k b1 ? 4.0 in.From Eq. (5): d ? 0.010 in.;L0 ? LaDerive a formula for the shortening d of the post due to theSquare cross sections:SHORTENING OF ELEMENT dyPdy. EAyEaB 1.5bSolution 2.3-14 Tapered post. SECTION 2.3 Changes in Lengths under Nonuniform Conditions 153.


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