材料力学1_物理必修1力学

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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 cro-sectional areas of members BD, BE, and CE of the tru shown.The working strees are 20 000 psi in tension and 12 000 psi in compreion.(A reduced stre in compreion is specified to reduce the danger of buckling.)Solution

The free-body diagram of homogeneous BC in Fig.(b).The equilibrium equation are MF0,163624PAy0,P=24(kips)=24000(lb)

The free-body diagram of tru in Fig.(c).The equilibrium equation are ME0,16PAy3(846416886416)PBD8360,PBD=－8.944(kips)(Compreion)

MB0,8PAy8PCE0

PCE=24(kips)

(Tension)

46416PBD0 Fy0,PAy0.707PBE36

PBE=-11.32(kN)(Compreion)

The normal stre of a member CE, DE and DF is

BDPBDABD8944lbABD12000(lb/in.)

2(Compreion)

ABD=0.745(in.2)BEPBEABE11320lbABE12000(lb/in.)

2(Compreion)

ABE=0.943(in.2)CEPCEACE24000lbACE20000(psi)

(Tension)

ABE=1.2(in.2).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 stre is 4000 psi.The cable attached at C has a diameter of 1/8 in.and a working normal stre of 20 000 psi.Solution The free-body diagram of bracket in Fig.(b).The equilibrium equation are

MFFxB0,6P2Tsin10PBxTcos100,P=0.05788T

(a)

PBx=-0.9848T 0,0,0,

PBy=0.2315T According to the normal strees formula, we have yPByPTsin104T0, 0.125in.2220000(lb/in.)T=245.4(lb)

2According to the shear strees formula, we have

VA4PBxPBy2

T=194.1(lb)According to formula(a), we get

P=0.05788T=0.05788×194.1=11.2(lb)

第二题（2010年3月10日前交该题作业）星期三 d24T0.98480.23152220.25in.4000(lb/in.)

2.37 An initially rectangular element of a material is deformed into the shape shown in the figure.Find εx, εy, and γ for the element.Solution According to the definition of the axial strain, we have:

xy

0.0040.20.15150.150.01

0.150.19920.2

16.615.70.90.0157

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.Solution Equilibrium: PABcos40PACPADcos20

1010(lb)

3PABsin40PADsin20

Compatibility

ycos40xsin40ABycos20xsin20ACy AD

PAC1012(in.)20106Hooke’s law: ACAB1010(lb/in.)0.6(in.)PAB1012(in.)/cos40622622PAC(in./lb)

62910(lb/in.)0.3(in.)PAD1012(in.)/cos206218.0110PAB(in./lb)PAD(in./lb)AD2910(lb/in.)0.3(in.)214.68106

we get 20cos40PACxsin4020cos20PACxsin2018.01PAB14.68PAD

we obtain

17.32PACand 6.160PAB9.436PAD

PABcos40PACPADcos201010(lb)

3PABsin40PADsin20

PAB0.5321PAD

PAD1.362PAC 17.32PAC6.160PAB9.436PAB0.532123.89PAB

PAB0.7249PAC 17.32PAC6.1600.5321PAD9.436PAD12.71PADSo we obtain PAC=3.53×103(lb)PAB=2.56×103(lb)PAD=4.80×103(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 stre developed in each material.The moduli of rigidity for steel, aluminum, and bronze are 12 x 106 psi, 4 x 106 psi, and 6 x 106 psi, respectively.Solution According to the angle of twist formula, we have

32LAC(TB750)GStedSte4

32LCD(TB500)GAludAlu4432LDBTBGBrodBro40

6(TB750)12144(TB500)423TB6140

0.5（TB-750）+0.0625(TB-500)+0.5TB=0 TB=382.4(lb.ft)According to the torsion formula, we have







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 maxSte16(382.4750)121322466(psi)898.4(psi)

16(382.4500)12maxAlu2316382.412maxSte1323370(psi)

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 compreive bending strees in the beam shown.Solution：

1）FBD(support reactions at A and B)

2）Shear-Moment Diagrams 3）Section Modulus Moment of inertia I＝100×106(mm)4 4）Maximum Bending Stre

At the top of section C it is in compreion

cMCctopIMCcbotIMBctopIMBcbotIMCctopIMBcbotI12.510N.mm130mm10010mm6646＝16.25（Mpa）

At the bottom of section C it is in tension

t12.510N.mm200mm10010mm664＝25（Mpa）

At the top of section B it is in tension

c1210N.mm130mm10010mm664＝15.6（Mpa）

At the bottom of section C it is in compreion

t1210N.mm200mm10010mm664＝24（Mpa）

So we get

c

t12.510N.mm130mm10010mm66464＝16.25（Mpa）

1210N.mm200mm10010mm＝24（Mpa）

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 stre at 1.0-in.vertical intervals on the cro section that carries the maximum shear force.Plot the results.Solution：

1)FBD(support reactions at B and C)

2)Shear-Moment Diagrams 3)Section Modulus Moment of inertia I＝97.0in.4

The first moment of this area at 1.0-in.vertical intervals on the cro section

Q141312in3＝12 in.3 or Q4132.511.2515.125in3

=15.125 in3 4)Maximum shear Stre

maxVmaxQIb225015.125in.97in.1.0in.343350.8psi=350.8psi Shear Stre at 1.0-in.vertical intervals on the cro section VmaxQ1Ib225012in.497in.1.0in.278.4psi=278.4psi

第六题（2010年4月12日前交该题作业）星期一

6.72 Compute the value of EIδ at the overhanging end A of the beam, by superposition.According to deflection formulas for beams, we know BwBCwBCa2

224LEI(a2L)2400(N/m)(2m)24(4m)EI32(224)m22600(N.m)EI

AMBCBwBC2m1200(N.m)EI BMBMB(2a)3EI4wABa8EI2MBaAwAB3EI4400(N/m)(2m)8EI

AMBBMB2m800(N.m)EI33800(N.m)(4m)3EI2m2133(N.m)EI3

AAwBCAMBAwAB1733(N.m)EI

7.44 The beam ABCD has four equally spaced supports.Find all the support reactions.Solution According to slope formulas for beams, we know B

w0(L)24EI[(3L)2(3L)(L)(L)]

323222[(3L)(L)(2L)]

[(3L)(L)(L)]0222RB(2L)(L)6(3L)EIRC(L)(L)6(3L)EI3

Cw0(2L)24EIRB(2L)6(3L)EIRC(L)(2L)6(3L)EI[(3L)2(3L)(2L)(2L)]

323[3L2L(2LL)2L[(3L)(2L)](2L)]

322222[(3L)(2L)(L)]0

22w0L2422w0L8RB187RB187RC188RC180 0

RBRC2433w0L30

According to the equilibrium equation and according to symmetry，A、D supports have the same magnitude，RARBRCRDw0(3L)

RARD we have

RBRC33w0L30,RARD2w0L5

第七题（2010年4月16日前交该题作业）星期五

8.27 The cro sections of the members of the pin-jointed structure are 200-mm square.Find the maximum compreive stre in member BDE.Solution The maximum compreive stre in member BDE

maxC

69010000.20.2258.110000.20.268.95(MPa)

第八题（2010年4月26日前交该题作业）星期一

8.49 For the state of stre shown, determine the principal strees and the principal directions.Show the results on a sketch of an element aligned with the principal directions.Solution

The principal strees 2

1xyxy2222xy

2

46246228

which yields

110.4(ksi)

28.4(ksi)The principal directions

tan22xy288

xy4651.6The two solution are

257.99

and 57.99180122.01

29.0

and 61.0

Determine the angles 1(aociated with 1)(aociated with 2)

xyxyx'22cos2xysin2

4646cos(229)8sin(229

22)

8.4(ksi)

and

and the angles 2



y'x246yxy2462cos2xysin2



Therefore we conclude the angles 1=61.0o(aociated with angles 229(aociated with 2)

210.4(ksi)

cos(229)8sin(229)

1)and the

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 stre is τw = 80 MPa, determine the smallest allowable diameter of the shaft.Neglect the weights of the pulleys and the shaft as well as the stre due to the transverse 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 have Mmax=2.5kN×0.6m=1.5(kN.m)TBC=0.3(kN.m)Therefore, the stre at the bottom of the section are

Mmax

S32Mmax3d321.510(N.mm)6d(mm)633 TBCrJ16TBCd3160.310(N.mm)d(mm)33Draw the Mohr’s circle

Append(2)

Figure(a)shows a reinforced concrete beam, where the cro-sectional area of the steel reinforcement is 19600 mm2.Using n = Est/Eco= 8 and the working strees of 12 MPa for concrete and 140 MPa for steel, determine the largest bending moment that the beam can carry.Solution

Suppose the tensile stre is zero and compreive stre is not zero in the concrete, the first moment of the transformed cro section about the neutral axis is zero.According to the maximum compreive stre in concrete and tensile stre in steel, the neutral axis is

hb1h2(b1b2)(hh1)22nAst(dh)0

(b)

9820h22(9820900)(h280)22819600(1670h)0

450h22654400h6115200000

h25898.7h13589330

h222.0(mm)

The moment of inertia about the neutral axis

IIb1h333(b1b2)(hh1)33nAst(h1h2h)22

(c)98202223819600(1670222)94

I(35.81328.8)10(mm)

364.610(mm)94

12(MPa)The moment

st364.610M19708(kN.m)

M(1670222)9comax222M9

The beam carries a uniformly distributed load of intensity wo on a simply supported span 24 ft long.Determine the largest allowable value of wo

M18w0l1828

364.610M4406.4(kN.m)

140(MPa)

24406.4w0(30)

w039.16(kN/m)

w04.00(t/m)

第十题（2010年5月7日前交该题作业）星期五

Append The 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.ypSolution Determine the moment of inertia of the cro-sectional area about the z-axis and y-axis IzIy1600mm(2400mm)122400mm(1600mm)1233

1.8432100.81921012(mm)1.8432(m)(imm)0.8192444

The slenderne ratio of a 1600mm x 2400mm rectangle The least radius of gyration with z-axis

rzIzA1.8432(m)421.62.4(m)37.5

0.48(m)

CzC29m0.48mThe least radius of gyration with y-axis

ryIyA0.8192(m)421.62.4(m)29.5

0.2133(m)

CyC0.79m0.2133mThe slenderne ratio

CC2E22225100020yCzC157

ypFor the slenderne ratio C,C is le than CC, so that the concrete column is of intermediate length.These equations yield the factor of safety

Nz53533CzC8CC3CyC8CCCzC8CCCyC33335353337.58157329.5815737.533815729.531.755 Ny8CC815731.736and the working stre

zw2CzC122CC2yp37.520111.07(MPa)2N1.7552157yw2CyC122CC2yp29.520111.32(MPa)2N1.7362157The largest allowable axial load thus becomes

PA11.07101.62.442.51210(N)42512(kN)4338(t)

PyywA11.321061.62.443.459106(N)43459(kN)4435(t)So we obtain axial load P

PP4338(t)

66zzwz

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