1、00 习题要求：1 将题目转化为 Microsoft Word 文档；2 解题过程用 Microsoft Word 文档，公式用公式编辑器，只交电子文档作业；3 期末考试前必须作对所有所给题目，否则不能参加期末考试，请于指定时间前交作业。第一题（2010 年 3 月 5 日前交该题作业）星期五1.21 Determine the smallest allowable cross-sectional areas of members BD, BE, and CE of the truss shown. The working stresses are 20 000 psi in tension
2、and 12 000 psi in compression. (A reduced stress in compression is specified to reduce the danger of buckling.)SolutionThe free-body diagram of homogeneous BC in Fig.(b). The equilibrium equation are, P=24(kips)=24000(lb)024361,0AyFPMThe free-body diagram of truss in Fig.(c). The equilibrium equatio
3、n are, 0368)16481648(316,0 BDAyE PPMPBD=8.944(kips) (Compression) PCE=24(kips) (Tension )0,CEAyB016437.,0 BDByyFPBE=-11.32(kN) (Compression) The normal stress of a member CE, DE and DF is (Compression)./(1208942inlbAlBDBDABD=0.745(in.2)(Compression)./(1032inlblPBEBEABE=0.943(in.2)(Tension )(04psilbP
4、CEECABE=1.2(in.2)1 .37 Compute the maximum force P that can be applied to the foot pedal. The l/4-in.-diameter pin at B is in single shear, and its working shear stress is 4000 psi. The cable attached at C has a diameter of 1/8 in. and a working normal stress of 20 000 psi.SolutionThe free-body diag
5、ram of bracket in Fig.(b). The equilibrium equation are, P=0.05788T (a)01sin26,0TPMB, PBx=-0.9848TcoFxx, PBy=0.2315Tsi, ByyAccording to the normal stresses formula, we haveT=245.4(lb)./(20.1542inlbinTAccording to the shear stresses formula, we have)./(40.2531984. 222 inlbidPAVByx T=194.1(lb)Accordin
6、g to formula (a), we getP=0.05788T=0.05788194.1=11.2(lb)第二题（2010 年 3 月 10 日前交该题作业）星期三2.702.37 An initially rectangular element of a material is deformed into the shape shown in the figure. Find x, y, and for the element.SolutionAccording to the definition of the axial strain, we have:04.2.019x15y 57
7、7.6.2.69（g） The bars AB, AC and AD are pinned together as shown in the figure. Calculate the axial force in the strut caused by the 10-kip load. For each steel bar. A = 0.3 in.2 and E = 29 x 106 psi. For the aluminum bar, A = 0.6 in.2 and E = 10 x 106 psi.SolutionEquilibrium: )(102cos40cos 3lbPPADCA
8、B 0ininCompatibilityABxy ssD2i20coyACHookes law: )/.(102.)(60)./(10626 lbinPinilbPACAC /8.3./294cos2 liiil BBA )/.(10)(0)./(1626 lbininilbADD we getABxACPP.84s40cos2D6iwe obtainAABAC3.916.3.17and )(102cos40cos 3lbPPDCB 0ininAA ADABP52.ADADACPP436.9521.06.32.17AD71.2C5.0. BB B89.AAB4So we obtainPAC=3
9、.53103 (lb)PAB=2.56103 (lb)PAD=4.80103 (lb)第三题（2010 年 3 月 15 日前交该题作业）星期一3.27 The compound shaft, composed of steel, aluminum, and bronze segments, carries the two torques shown in the figure. If TC = 250 lb.ft, determine the maximum shear stress developed in each material. The moduli of rigidity for
10、 steel, aluminum, and bronze are 12 x 106 psi, 4 x 106 psi, and 6 x 106 psi, respectively.SolutionAccording to the angle of twist formula, we have032)50(32)750(32 444 BrorDAlulBCDStetBAC dGTLdTLdGTL16)(16444 0.5（T B-750）+0.0625(T B-500)+0.5TB=0TB=382.4(lb.ft)According to the torsion formula, we have
11、)(2461)7504.382(6max psiSte .893Alu)(70.max siSte第四题（2010 年 3 月 19 日前交该题作业的剪力图、弯矩图部分）星期五第四题（2010 年 3 月 24 日前交该题作业的应力部分）星期三5.35 Determine the maximum tensile and compressive bending stresses in the beam shown.Solution：1） FBD(support reactions at A and B)2） Shear-Moment Diagrams3） Section ModulusMomen
12、t of inertiaI100 106 (mm)44） Maximum Bending StressAt the top of section C it is in compression16.25 （Mpa）46103.5.2mNIcMtopc At the bottom of section C it is in tension25 （Mpa）462IbotCtAt the top of section B it is in tension 15.6 （Mpa）46103.2mNIctopc At the bottom of section C it is in compression2
13、4 （Mpa）462.IMbotBtSo we get16.25 （Mpa）46103.5.2mNIctopCc 24 （Mpa）2.botBt第五题（2010 年 3 月 29 日前交该题作业）星期一5.68 For the beam shown, compute the shear stress at 1.0-in. vertical intervals on the cross section that carries the maximum shear force. Plot the results.Solution：1)FBD(support reactions at B and C
14、)2)Shear-Moment Diagrams3)Section ModulusMoment of inertiaI97.0in. 4The first moment of this area at 1.0-in. vertical intervals on the cross section12 in. 33112inQor 31255.4in=15.125 in34)Maximum shear Stress=350.8psipsiinibIQV8.350.1.97204maxa Shear Stress at 1.0-in. vertical intervals on the cross
15、 section=278.4psisiiiI .2.0.5431ax第六题（2010 年 4 月 12 日前交该题作业）星期一6.72 Compute the value of EI at the overhanging end A of the beam, by superposition.According to deflection formulas for beams, we knowEImNEImNLaEIwBC ).(60)42()4(2/0)(24 22ImAM.13IBB3)( IIBMA ).(2133)(.802EINEINawA ).(8)2(/40834mAwBMBC.
16、1737.44 The beam ABCD has four equally spaced supports. Find all the support reactions.SolutionAccording to slope formulas for beams, we know)()3(2)(24320 LLEIwB 36)(R 0)()3(622LEIRC)2()3(2)(2430IC )()3(6 3LLLEB 0)(2)(2IRC18740CBRw02L30RCBAccording to the equilibrium equation and according to symmet
17、ry，A、D supports have the same magnitude，)3(0LwRDCBDARwe have, 30LwCB 520LwRDA第七题（2010 年 4 月 16 日前交该题作业）星期五8.27 The cross sections of the members of the pin-jointed structure are 200-mm square. Find the maximum compressive stress in member BDE.SolutionThe maximum compressive stress in member BDE)(95.
18、682.01582.0196max MPaC第八题（2010 年 4 月 26 日前交该题作业）星期一8.49 For the state of stress shown, determine the principal stresses and the principal directions. Show the results on a sketch of an element aligned with the principal directions.Solution The principal stresses221 xyyxyx 2864which yields)(.10ksi)(.
19、2ksiThe principal directions6.15842tanyxThe two solution areand 9.572 01.289.57and 0061Determine the angles (associated with ) and the angles 2(associated with )22sinco xyyxyx )9(8)9s(64)(.8ksiand2sinco2 xyyxyxy)9(8)9s(64)(.10ksiTherefore we conclude the angles =61.0o (associated with ) and the 11an
20、gles (associated with )292第九题（2010 年 4 月 30 日前交该题作业）星期五8.108 A shaft carries the loads shown in the figure. If the working shear stress is w = 80 MPa, determine the smallest allowable diameter of the shaft. Neglect the weights of the pulleys and the shaft as well as the stress due to the transverse
21、shear force. Solution According the support, we know there is the largest bending moment occurs at C, the largest torque occurs in segment BC. Show in the figure.At C section, we haveMmax=2.5kN0.6m=1.5(kN.m)TBC=0.3(kN.m)Therefore, the stress at the bottom of the section are363maxax )(.105.2mdNdS3363
22、 )(16TJrBCBCDraw the Mohrs circleAppend(2)Figure (a) shows a reinforced concrete beam, where the cross-sectional area of the steel reinforcement is 19600 mm2. Using n = Est/Eco= 8 and the working stresses of 12 MPa for concrete and 140 MPa for steel, determine the largest bending moment that the bea
23、m can carry.SolutionSuppose the tensile stress is zero and compressive stress is not zero in the concrete, the first moment of the transformed cross section about the neutral axis is zero. According to the maximum compressive stress in concrete and tensile stress in steel, the neutral axis is(b)0)(2
24、)()(21211 hdnAhbhst)167(98098092655402 hh137.)(mThe moment of inertia about the neutral axis(c)21312131 )()(hnAhbhI st26709829049)(1).3.5(mI49(64The moment)(210.329maxMPaco)(978kNM)(1406.34)217(89MPaMst 0mkNThe beam carries a uniformly distributed load of intensity wo on a simply supported span 24 f
25、t long. Determine the largest allowable value of wo2081lM2)3(4.6/90mkN.tw第十题（2010 年 5 月 7 日前交该题作业）星期五AppendThe 9-m-long concrete column is built in at its base and stayed by two beam at the top. Determine the largest axial load that can be carried. Use E =25 GPa and =20 MPa for concrete.ypSolutionDe
26、termine the moment of inertia of the cross-sectional area about the z-axis and y-axis 44123 )(832.)(084.12)40(6 mmIz 19.09.iyThe slenderness ratio of a 1600mm x 2400mm rectangleThe least radius of gyration with z-axis)(48.0)(.261843mAIrzz 57.09mCzThe least radius of gyration with y-axis)(213.0)(4.61
27、8AIryy 5923.07CyThe slenderness ratio157202ypCEFor the slenderness ratio is less than , so that the concrete zCy, Ccolumn is of intermediate length. These equations yield the factor of safety75.18.3157.3835 CzzzN629.3yyand the working stress)(07.15.172.321 MPaNCypzzw 32.69ypyyThe largest allowable axial load thus becomes)(438)(451)(1052.4.6107. tkNAPzw 53993326ySo we obtain axial load P)(48tz
