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gere goodno 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. 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 crosssectional 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 crosssectional 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\nstressstrain 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\nstressstrain 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 1.67\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 crosssectional 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 secondfloor 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 234 (24) 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 crosssectional area of the middle bar (material B ) is 50% larger than the\ncrosssectional area of one of the outer bars. The end segments have crosssectional\narea A 1 840 mm\n \n 2 and length L 1 200 mm. The middle segment has\ncrosssectional 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 crosssectional 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 crosssectional area A 3 supports the bar at B.Both wires have the same crosssectional 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 precompress the CI pipe. The pitch of the threads of the\nbolt is p 52 mils (a mil is onethousandth 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. Highstrength 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 posttensioned 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 crosssectional 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 recentering\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\ncrosssectional 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 2.72 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 crosssectional 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 highstrength 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 stressstrain 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 stressstrain curve that\nmay be idealized as elastoplastic with yield stress Y. Each bar has \ncrosssectional area A.\n \n Determine the yield load PY and the plastic load PP.\n \n Problem 2.122 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. ThinWalled Tubes 328. Stress Concentrations in Torsion 338ShearForce and BendingMoment 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. BuiltUp Beams 466. Beams with Axial Loads 475. Stress Concentrations 492TransformedSection Method 508. Beams with Inclined Loads 520. Bending of Unsymmetric Beams 529. Shear Stresses in WideFlange Beams 541. Shear Centers of ThinWalled 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 BendingMoment Equation 714. Deflections by Integration of the Shear Force and Load Equations 722. Method of Superposition 730. MomentArea 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. ParallelAxis 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 crosssectional 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 crossThe cables are combined at point Q,Each cable has an effective crossTable H1 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 crosssectional area of the cableThe crane boom liesAeFor materials in tension, we may use a characteristic tensile stress (as obtained from a stressstrain curve) as a measure ofNote that the ratio has units of length.
Using the ultimate stress U as the strength parameter, calculate the strengthtoweight ratio (in units of meters) forAluminum is higher than steel, which makes it desirable forThe angle between the inclined barsThe test specimen Prob. 1.33). 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 stressThe 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 modWhat 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 memThe 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 diamPaT1 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 crosssectional area and total weight W. A vertical spring of stiffness k supports the arm at point B.

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