
gere mechanics of materials 8th edition solution manual
gere mechanics of materials 8th edition solution manual
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gere mechanics of materials 8th edition solution manual
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25 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 2.24 Cage supported by a cableL1 ? 4.6 m. L2 ? 10.5 m. EA ? 10,700 kN. W ? 22 kNProblem 2.26 The device shown in the figure consists of a prismatic rigid pointer ABC supported by a uniform translaInclude 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 2.25 Safety valveAll Rights Reserved. Solution 2.26Sum 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 2.27 Two rigid bars are connected to each other by two linearly elastic springs. Before loads are applied, theRigid barRigid barAll Rights Reserved. Solution 2.27Cut 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 2.28 The threebar truss ABC shown in figure part a has a span L ?
3 m and is constructed of steel pipes havingAz. Ay. Ax. Bz. BySolution 2.28E ? 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 ydirection at joint A. Sum forces in ydirection 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 2.210 A uniform bar AB of weight W ? 25 N is supported byThe spring on the left has stiffnessNeglect the weight of the springs.Solution 2.29Load 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 2.21 The Lshaped 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 2.22 A steel cable with nominal diameterSolution 2.22 Bridge section lifted by a cable. All Rights Reserved. A ? 304 mm2Table 21). W ? 38 kN. E ? 140 GPa. L ? 14 mPULT ? 406 kN (from Table 21). Pmax ? 70 kNThe moduli of elasticity for the steel andSolution 2.23EsLa. LsEsEa APLs. Es AEa APLs. Es AEaAsEaEaEa. Problem 2.24 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.
211 A hollow, circular, castiron pipe (Ec ? 12,000 ksi)W and its own weight?Steel cap. Cast iron pipeBrass rodSolution 2.211PIPE, 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 I1). 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 2.212 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 2.213 Two pipe columns (AB, FC) are pinconnected to a rigid beam (BCD) as shown in the figure. Each pipeAssume the innerBCD displaces downward to a horizontal posiBCD displaces downward to a horizontal posiRigid 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 2.214. Apply the laws of statics to the structure in its displaced position; also use FBD’s of the left and right bars aloneProblem 2.214 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.
215 Solve the preceding problem for the following data. Solution 2.215. 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 2.31Solution 2.31P ? 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 2.32 A long, rectangular copper bar under a tensile load PEs ? 200 GPa.CopperProblem 2.33 An aluminum bar AD (see figure) has a crossP2 ? 1200 lb, and P3 ? 1300 lb. The lengths of the segments of theDoes the bar elonSolution 2.33So 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 2.34 A rectangular bar of length L has a slot in theThe bar has width b,Assume that the axial stress in the middle regionSolution 2.34Lslot ?LslotLslotE ? 210 GPa L ? 750 mm smid ? 160 MPaAll Rights Reserved. Problem 2.35 Solve the preceding problem if the axial stressIn part (c), assume that. Solution 2.35Problem 2.36 A twostory building has steel columns AB in the first floorThe roof load P1 equalsEach column has length. L ? 3.75 m. The crosssectional areas of the first and secondfloor columnsLslot ?LslotLslotSo smid ?E ? 30,000 ksi L ? 30 in. smid ? 24 ksiAll Rights Reserved. Problem 2.37 A steel bar 8.0 ft long has a circular cross section of diameter in.The modulus of elasticity psi.P ? 5000 lb?Solution 2.36 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 2.38 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 2.39 A wood pile, driven into the earth, supports aThe friction force f per unit length of pile is assumed to beThe pile hasSolution 2.39N(L) ? f s(y) ?N(y)SkinSkinSkinSkin friction fCompressive stressUse the properties and dimensionsSolder joints. Tinlead 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 2.310. Problem 2.311 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 2.312 A prismatic bar AB of length L, crosssectional area A, modulus of elasticity E, andSee Appendix I for weight densities of steel and seaA ? 0.0157 m2All Rights Reserved. Solution 2.312 Prismatic bar hanging verticallyIn sea water. In air. W ? (gs)AL ? 1813.35kN d ?E ? 210 GPaConsider an element at disWydyWyN(y)dyW ?
Weight of bar (b) ELONGATION OF BARElongation of upper half of bar. Elongation of lower half of bar:All Rights Reserved. Problem 2.313 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 2.314 Tapered post. SECTION 2.3 Changes in Lengths under Nonuniform Conditions 153.

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