2009机制材料力学试卷A
一、（20）
The solid 80-mm-diameter bar carries a torque T and a tensile force of 125 kN acting 20 mm from the centerline of the bar. Find the largest safe value of T if the working stresses are σw =100 MPa and τw = 80 MPa.
承受偏心距为20mm，偏心拉力为125kN，扭矩T的直径为80-mm 的圆截面杆，其许用正应力为σw =100 MPa，许用剪应力为τw = 80 MPa求最大扭矩。
Solution
Mmax=125kN×0.02m=2.5(kN.m) MP32MmaxP??max???3SA?dA 7 6332?2.5?10(N.mm)4?125?10(N)??33??80(mm)??802(mm)2 ＝49.7＋24.9＝74.6 （Mpa）
16?T?106(N.mm)????＝9.95T 6 3J?d??803(mm)3The unit of T is (kN.m)
Draw the Mohr’s circle
Tr16TAccording to Mohr’s circle, we have
???max??()2??2??W
222Settle it, we get
??max?()2??2??W
?max?(37.3?37.32?(9.95T)2)?100MPa T=5.07(kN.m)
?max?(37.32?(9.95T)2)?80MPa 6 T=7.11(kN.m) So we get
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T=5.07(kN.m) 二、（20）
The timber beam is reinforced with steel plates rigidly attached at the top and bottom. The allowable stresses are 8 MPa for wood and 120 MPa for steel, and the ratio of the elastic moduli is Est/Ewd = 15. Determine the increase in the allowable bending moment due to the reinforcement.
上下用钢板加固的木梁，弹性模量比为Est/Ewd = 15，木头许用应力为 8 MPa ，钢板许用应力为120 MPa 。求加固的木梁比未加固的木梁增加的弯矩。
Solution
As a result of Est/Ewd = 15，the results in the transformed section shown in Fig.(b), which is made of wood.
nAst?15?120?10?18000(mm2)
nb?15?120?1800(mm)
The moment of inertia 5
1800?3203(1800?250)?3003I???(4.915?3.488)?109?1.43?109(mm4)
1212
The largest bending stress
McwdM?150??8(MPa) 5 I1.43?109 M?76.3?106(N.mm)?76.3(kN.m)
McM?160 ?st?nst?15??120(MPa) 5 9I1.43?10 M?71.5?106(N.mm)?71.5(kN.m)
?wd?So we get M?71.5?106(N.mm)?71.5(kN.m)
If there only is the wood, the moment of inertia
250?3003I??0.5625?109(mm4)
12The largest bending stress
2
McwdM?150??8(MPa) 9I0.5625?10 M?30.0?106(N.mm)?30.0(kN.m)
?wd?So increase in the allowable bending moment
71.5?30.0?41.5(kN.m) 5
三、（20）
The propped cantilever beam carries two concentrated loads, each of magnitude P. Find the support reaction at A and B。 求支座的反力。
Solution
According to the slope and displacement formulas for beams, we know
?2L??L?P???P???RA?L32LL3?3????A??(3L?)?(3L?)??0
6EI36EI33EIRAL328PL37PL3?A????0 10
162EI162EI3EI22
We have
63P 2 162According to the equilibrium equation，we have RA?RB?0
2PLPL ?LRA??MB??0
33963MB?PL ， RB??P
162162So we have
96363MB?PL ， RA?P ， RB??P 8
162162162四、（20）
RA?The beam shown in cross section is fabricated by bolting three 80-mm by 200-mm wood planks together. The beam is loaded so that the maximum shear stress in the wood is 1.4 Mpa. If the maximum allowable shear force in a bolt is Fw=8kN, determine the largest permissible spacing of the bolts.
如图所示结构梁，由三块80-mm X 200-mm的木块用螺钉固定在一起，已知：木头的最大剪应力为1.4 Mpa，螺钉的许用剪力为Fw=8kN,
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求沿木梁方向螺钉的最大间距（垂直纸面）。 Solution
(a)The shear stress at the neutral axis
Q?200?80?140?80?100?50?2.248?106(mm)3200?3603120?2003 I???697.6?106(mm)4
1212
?maxVmaxQV?2.248?106?5???4.028?10V?1.4 6Ib697.6?10?80 V?34756(N) 7
(b) The shear stress at 100mm above the neutral axis.
Q?200?80?140?2.240?106(mm)3
200?3603120?2003I???697.6?106(mm)4
1212VQ34756?2.24?106??1.395(MPa) 7 ??6Ib697.6?10?80The largest permissible spacing of the bolts 1.395(MPa)???五、（20）
Fw8?1000 ?80L80L L?71.68(mm) 6
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 ?yp=20 MPa for concrete.
五、（20）如图所示9m高的混凝土柱，E = 25 GPa ， σyp = 20 MPa，求最大载荷P。
Solution
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Determine the moment of inertia of the cross-sectional area about the z-axis and y-axis
1600mm?(2400mm)3Iz??1.8432?1012(mm)4?1.8432(m)4
122400mm?(1600mm)3Iy??0.8192?102(imm)4?0.8192 4
12The slenderness ratio of a 1600mm x 2400mm rectangle The least radius of gyration with z-axis rz?Iz1.8432(m)4??0.48(m) 2A1.6?2.4(m) CzC?2?9m?37.5 4 0.48mThe least radius of gyration with y-axis ry?Iy0.8192(m)4??0.2133(m) A1.6?2.4(m)2 CyC? CC?0.7?9m?29.5 4
0.2133m2?2E?2?2?25?1000?157 4
20The slenderness ratio
?ypFor the slenderness ratio CyC,CzC is less than CC, so that the concrete column is of intermediate length. These equations yield the factor of safety
53CzCCzC53?37.537.53?????1.755 Nz??3338CC8CC38?1578?15753CyCCyC53?29.529.53?????1.736 Ny??38CC8CC338?1578?157333and the working stress ?zw ?yw?CzC2??yp?37.52?20??1??1???11.07(MPa) ??22??2CC??N?2?157?1.755??CyC2??yp?29.52?20??1??1???11.32(MPa) 3 ??22???2?157?1.736?2CC??NThe largest allowable axial load thus becomes
Pz??zwA?11.07?106?1.6?2.4?42.512?106(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?Pz?4338(t) 1
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