MAE_101-Midterm-Solution
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MAE_101-Midterm-Solution

Course Number: MAE 101, Spring 2008

College/University: UCLA

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Midterm - Solution 1 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 1 The horizontal bar of weight W is supported by a roller support at A and the cable BC. Apply rigorous analysis of equilibrium to determine the angle , the tension in the cable, and the magnitude of the reaction at A. (The answer may be intuitively obvious, but prove it using basic principles) Free body diagram and...

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- Midterm Solution 1 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 1 The horizontal bar of weight W is supported by a roller support at A and the cable BC. Apply rigorous analysis of equilibrium to determine the angle , the tension in the cable, and the magnitude of the reaction at A. (The answer may be intuitively obvious, but prove it using basic principles) Free body diagram and equations of equilibrium Ay T M ( B) Fx Fy LAy T cos Ay W W L W 0 2 0 (*) T sin W 2 Ay W 2 W (**) 2 T sin 0 T sin From (*) either T=0 or cos =0 If T=0 then (**) cannot be satisfied If cos =0 then =/2 and T=W/2 MAE 101. Statics and Strength of Materials G. E. Orient 2006 2 Problem 2 The Pratt bridge truss supports loads at F, G, and H. Determine the axial forces in members BC, BG, and FG. l=4m, F1=60 kN, F2=80 kN, F3=20 kN Global FBD, Equilibrium =45o l Ay Ax F1 M ( A) Fy Ay Fx Ax F1l F1 0 Ey l F2 F2 2l F2 F3 Ax F3 3l Ey 0 F1 l F2 l F3 l F3 E y 4l 0 0 Ay Ey F1 F1 2 F2 3F3 70 kN 4 F2 F3 E y 90 kN Method of joints Joint A Ay PAB PAF Fy Fx Ay PAB sin PAF 0 0 PAB PAF Ay sin PAB cos Ay 2 127.3 kN Ay tan 90 kN PAB cos 3 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 2 cont'd Method of joints Joint F PFB PAF PFG Fx Fy PAF PBF PFG F1 0 0 PFG PBF F1 PAF 90 kN (T ) 60 kN F1 Joint B /2- PBA PBF Fy Fx PBA sin PAB cos PBF PBC PBG sin PBG cos 0 0 PBG PBC PBF sin ( PAB PAB PBG ) cos Ay sin ( 2 Ay F1 ( Ay F1 ) 2 42.43 kN (T ) 120 kN (C ) PBC PBG F1 ) cot 4 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 3 What is the axial load in member BD of the frame? l1=10 m, l2=5 m, l3=5 m, w=100 N/m w Replace distributed load with statically equivalent forces, FBD and global equilibrium l3 F3 l2+l3/2 l2 F2 Ay Ey 2l2/3 Ex Magnitudes of statically equivalent loads F2 wl 2 / 2, F3 wl 3 Equilibrium l 2l M (E) Ay l1 F2 2 F3 l2 3 0 3 2 l1 Ay F2 2 l2 3 l1 F3 l2 l1 1 l3 2 l1 458.33 N 5 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 3 cont'd FBD, equilibrium of frame member ABC: Cy Cx Bx Ay BD is two-force member M (C ) Bxl3 Ay l1 0 Bx Ay l1 l3 916.67 N (T ) 6 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 4 The mass of the van is m=2250 kg, and the coefficient of static friction between its tires and the road is s=0.6. If its front wheels are locked and its rear wheels can turn freely, what is the largest value of for which it can remain in equilibrium? What is the largest value of for which the van can remain in equilibrium if it points up the slope? Discuss the effect of weight on the results. Geometry: l1=1.2 m, l2=3 m, l3=1 m (a) FBD and equilibrium R N N MR Fx (l2 l1 )W cos s N W sin W l3 l3W sin 0 l2 W sin s l2 N 0 l1 l2 Doesn't depend on weight (l2 l1 )W cos tan l2 l1 l2 l3 s l3W sin 0 l1 24.2o 7 1 / l2 s 1 s l3 / l2 0.45, MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 4 cont'd (b) FBD and equilibrium N W R N MR Fx (l2 l1 )W cos sN l3W sin 0 l2 W sin s l2 N 0 W sin (l2 l1 )W cos tan l2 l1 l2 l3 s l3W sin 1 l1 / l2 s 1 s l3 / l2 0 16.7 o 0.3, 8 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 5 The cross-sectional area of bar AB is 0.015 m2. If the force F = 20 kN, what is the normal stress on a plane perpendicular to the axis of bar AB? The bar AB consists of a material that will safely support a tensile normal stress of max=20 MPa. Based on this criterion, what is the largest safe value of the force F? l1 l2 (a) FBD, equilibrium equations, average stress Cx Cy PAB PAB F l1 l2 l1 sin M (C ) AB F (l1 l2 ) PAB sin l1 F l1 l2 AAB l1 sin 0 PAB F 34641 N PAB (b) Safe load Maximum AB member force PAB, max PAB, max AB , max AB , max AAB AAB Moment equilibrium for maximum AB member force M (C ) Fmax (l1 l2 ) PAB, max sin l1 0 Fmax PAB, max sin l1 l1 l2 AB, max PAB AAB 2.31 MPa AAB sin l1 l1 l2 9 173.2 kN MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 6 Each bar has a cross-sectional area of 3 in2 and modulus of elasticity E = 12106 lb/in2. If a 40-kip horizontal force is applied at A, what are the normal stresses in the bars? Show derivations related to displacements. If a large system of equations emerges (>2x2) you don't have to solve it. Answer the question symbolically in terms of notations you introduced Geometry l l l l AB , l AD , l AC sin sin sin FBD of joint A, equilibrium l PAD - PAB F P AC Fy PAB sin PAD sin PAC sin 0 Fx F PAB cos PAD cos PAC cos 0 Force-deformation; introduce stiffness parameters kAB, kAD, kAC PABl AB PABl k AB PAB , AB AE AE sin PAC l AC PAC l k AC PAC , AC AE AE sin PADl AD PADl MAE 101. Statics and Strength of Materials k AD PAD AD G. E. Orient 2006 AE AE sin 10 Problem 6 cont'd Displacement compatibility derivations for small deformations (u20, v20, 20) B D C A' Member AB l AB cos u l AB cos2 2 A 2 u 2 AB v v2 2 2 AB AB l AB sin v 2 l AB 2 2ulAB cos u 2 l AB sin 2 2vl AB sin l AB 2l AB u cos v sin AB Similarly, member AD: u cos v sin AD Member AC, however is different: l AC cos u cos u 2 l AC sin AC v 2 l AC 2 AC v sin 11 MAE 101. Statics and Strength of Materials G. E. Orient 2006 Problem 6 cont'd Displacement compatibility AB AD AC k AB PAB k AD PAD k AC PAC u cos u cos u cos v sin v sin v sin System of equations for five unknowns s1 c1 k AB 0 0 s2 c2 0 k AD 0 s3 c3 0 0 k AC 0 0 c1 c2 c3 0 0 s1 s2 s3 PAB PAD PAC u v 0 F 0 0 0 P_AB=22344,P_AD=9581,P_AC=30502,u=0.087,v=-0.0136 Stresses in truss members PAB / A 7.45 ksi AB AD AC PAD / A 3.19 ksi PAC / A 10.17 ksi 12 MAE 101. Statics and Strength of Materials G. E. Orient 2006

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