Name (Print) ME Mechanics of Materials Exam # 3 Date: December 9, 2013 Time: 7:00 9:00 PM Location: EE 129 & EE170

Transcription

1 Name (Print) (Last) (First) Instructions: ME Mechanics of Materials Exam # 3 Date: December 9, 2013 Time: 7:00 9:00 PM Location: EE 129 & EE170 Circle your lecturer s name and your class meeting time. Krousgrill Sadeghi Bilal 7:30-8:30AM 10:30-11:30AM 3:30-4:30PM Begin each problem in the space provided on the examination sheets. If additional space is required, use the yello paper provided. Work on one side of each sheet only, ith only one problem on a sheet. Please remember that for you to obtain maximum credit for a problem, you must present your solution clearly. Accordingly, coordinate systems must be clearly identified, free body diagrams must be shon, units must be stated, rite don clarifying remarks, state your assumptions, etc. If your solution cannot be folloed, it ill be assumed that it is in error. When handing in the test, make sure that ALL SHEETS are in the correct sequential order. Remove the staple and restaple, if necessary. Prob. 1 (24 points) Prob. 2 (8 points) Prob. 3 (30 points) Prob. 4 (8 Points) Prob. 5 (26 points) Prob. 6 (4 points) Total

2 ME 323 Examination #3 December 9, 2013 Name (Print) (Last) (First) Instructor Useful Equations (If you do not see equations that you need, raise your hand and ask.) E E G G 21 P PABL ub ua AB AB TL A AE T TABL B A AB AB I GI P dv dm d x ( ) V EI M dx dx dx dv V P M M dx My VQ Q Ay I It x y x avg cos2 xy sin 2 2 x y y avg cos2 xy sin 2 2 x y x y sin 2 xycos2 2 R R tan 2 1 avg 2 avg P avg x y x y R 2 2 n L 0 L 2 2 xy n 0,1,2,... 0 n1 0 xa n xa dx for x a x a for n n x a for x a n1 x a for n 0 n F x 1 F L 1 T x 1 T L U dx U 2 EA 2 EA 2 dx GI 2 GI x xy 0 P 2 L 2 s p 0 y p 0 / M x 1 f V x U dx U dx 2 EI 2 GA U i P i M( x) F( x) T( x) V( x) L M( x) L F( x) LT( x) L fsv( x) Pi Pi Pi P i i dx dx dx dx EI EA GI AG L P Page 2 of 12

3 ME 323 Examination #3 December 9, 2013 Name (Print) (Last) (First) Instructor h pr t a pr 2t 1 x x y z T E 1 y y x z T E 1 z z x y T E G G G xy xy xz xz yz yz 2 M failure stress yield strength FS, alloable stress state of stress M x y y z x z xy yz xz S Factor of safety based on the maximum shear stress theory y /2 maxabs 1/ d d bh IP Icircle Irectangle r Semi circle : y 3 Page 3 of 12

4

5

6 ME 323 Examination #3 December 9, 2013 Name (Print) (Last) (First) Instructor PROBLEM 2 (No partial credit. Zero credit if more than one choice is selected) 2a - (4 Points) - A beam is subject to the loading condition as shon belo. y W 0 M x M 0 R 1 L/3 L/3 L/3 R The loading function for the beam according to singularity/macaulay function is: (choose the appropriate loading function). a) ( x) R x xl/ 3 M xl/ 3 R xl M xl b) ( x) R x xl/ 3 M xl/ 3 R xl M xl c) ( x) R x xl/ 3 M xl/ 3 R xl d) ( x) R x xl/ 3 M xl/ 3 R xl M xl e) None of the above M xl The shear function for the beam according to singularity/macaulay function is: (choose the appropriate loading function). a) V( x) R x xl/ 3 M xl/ 3 R xl M xl b) V( x) R x xl/ 3 M xl/ 3 R xl M xl c) V( x) R x xl/ 3 M xl/ 3 R xl M x d) V( x) R x xl/ 3 M xl/ 3 R xl M xl e) None of the above L Page 6 of 12

7 ME 323 Examination #3 Name (Print) (Last) (First) December 9, 2013 Instructor 2.b - (4 Points) The state of stress at a point is given as shon belo. Y Choose hich Mohr s Circle is appropriate for the state of stress shon. The principle stresses are p1 p2 p3, and 40 ksi max is the maximum in-plane shear stress. 10 ksi 20 ksi X (c) (c) (1) max (2) max x, xy x, xy p3 p p1 p3 p p1 y, xy y, xy (c) (c) (3) max (4) p1 x, xy p y, xy p3 p1 x, xy p y, xy p3 max Page 7 of 12

8 ME 323 Examination #3 December 9, 2013 Name (Print) (Last) (First) Instructor Problem 3 (30 points): - The member is fixed into a all and subject to the loading as shon belo. 1. Dra the free body diagram for the member and determine the reaction forces and moments at the all. 2. Determine the state of stress at point A. 3. If the yield strength for the member is S y = 40 Ksi, determine the factors of safety guarding static failure at point A, based on the maximum shear stress and distortion energy theories. Note: P = 400 lb, L=10 in and member diameter = 2 in. Page 8 of 12

9

10

11

12

13

14

15 ME 323 Examination #3 December 9, 2013 Name (Print) (Last) (First) Instructor Problem 4 (8 points): - The beam is subject to the loading as shon in the figure belo. 1. Choose the correct shear diagram from the column on the left. 2. Choose the correct moment diagram from the column on the right. Note: The correct shear diagram on the left and the correct moment diagram on the right do not necessarily line up. Circle ansers from the folloing choices. Shear Diagram A) A) Bending Moment Diagram B) B) C) C) D) D) E) None of the above E) None of the above Page 10 of 12

16 Problem 5 (20 points): - To rods (each made up of a material ith a Young s modulus E, a thermal expansion coefficient α and circular cross-section of diameter d) are attached beteen a rigid disk C (of eight W) and to fixed supports B and H, as shon in the figure. Rod 1 has a length of L, hereas Rod 2 has a length of 2L. Rod 1 is given a temperature increase of ΔT ith the temperature of Rod 2 being held constant. Ignore the eights of the rods. H d 2 2L a) Dra a free body diagram of the rigid disk C and rite don the equilibrium equation(s) for the disk. rigid disk of eight W C F x = F 2 W F 1 = 0 F 2 = F 1 + W F 2 B d 1 L W + F 1 b) Write don the load/temperature/elongation equation for each rod. e 1 = F 1L EA + α ΔT L e 2 = F 2(2L) EA c) Write don the compatibility equation relating the elongation/compression of the to rods. u H = e 1 + e 2 = 0 d) Solve your equations from a), b) and c) above for the stress in Rod 1. Leave your anser in terms of, at most, W, ΔT, α, E, d and L. 0 = e 1 + e 2 = F 1 L EA + α ΔT L + 2F 2 L EA = F 1 L EA + α ΔT L + 2( F 1 + W ) L = 3F 1 L EA EA + 2WL EA + α ΔT L 2W + α ΔT EA F 1 = 3 σ 1 = F 1 A = 1 2W 3 A + α ΔT E = 1 8W 3 πd 2 + α ΔT E ( compression)

17 ME 323 Examination #3 December 9, 2013 Name (Print) (Last) (First) Instructor Problem 6 (4 points): - Consider to situations. One, here the beam belo is made from steel, having a Young s modulus of E st. To, here the beam is made from aluminum, having a Young s modulus of E al, here E st E al. Let max st and max al denote the absolute values of the maximum normal stress on the cross section at location B for the steel and aluminum beams, respectively. Circle the correct statement belo related to the relative sizes of max st and max al : a) max b) max c) max st max al st max al st max al Page 12 of 12