gere mechanics of materials solution manual
gere mechanics of materials solution manual
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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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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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