EM121 Statics and Mechanics of Materials I Concepts Learned for Exam 1 2D vectors Demonstrate the addition of vectors graphically and mathematically, using correct notation. 3D vectors Find the resultant of multiple 3D forces by adding like components.

Given a 3D vector, find the direction cosines. Given a 3D vector, find the unit vector in the direction of the vector. FBDs & 2D particle equilibrium Draw a FBD given a system boundary. Use a FBD with force equilibrium to solve particle equilibrium problems in 2D. FBDs & 2D particle equilibrium Replace commonly encountered 2D supports with their idealized reactions on a FBD.

Solve 2D equilibrium problems that require more than one system to find the desired values. FBDs & 3D particle equilibrium Solve 3D equilibrium problems by expressing forces as vectors and adding like components. Stress: tension, compression, & shear Identify the stressed areas for members under load. Find the normal and shear stresses of members under load. Stresses on inclined planes Calculate the normal and shear stresses at angled cuts of members in uniaxial loadings. Stress-strain relationships Identify important features of a stress-strain curve and identify material properties. Use Hooke’s Law to relate stress to strain. Strain: thermal & mechanical Calculate the strain or the change in length of a body subjected to a change in temperature, an axial load, or a combination. Concepts Learned for Exam 2 Statically indeterminate axial loading Use the force-deflection relationship or the temperature-deflection relationship to find loads in statically indeterminate systems. Factor of safety Calculate the factor of safety for a known load and material strength.

In a system with complex loading or geometry, find the overall factor of safety, maximum allowed load, or the minimum geometry. Moment about a point Calculate the moment about a point due to a force, using the vector or scalar approach. Couples Find the moment about a point due to a force couple, using the vector or scalar approach. Use the force couple formula to find the

distance between the force pair. Centroids by Integration Locate the centroid for a set of distributed point masses in 2D or 3D space.

Locate the centroid for an area in 2D space by setting up the pieces of the integral formula shown in class. Centroids by Composite Bodies Locate the centroid of a 2D shape by treating it as a collection of simple shapes, and document your work as we did in class. Distributed Loads Find the equivalent force and line of action that approximates a given distributed load on a horizontal beam. Concepts Learned for Exam 3 Drawing FBDs for rigid bodies Draw FBDs of 2D bodies where the reactions include moments in addition to forces. FBDs & 2D rigid body equilibrium Apply force and moment equilibrium equations to 2D rigid bodies to solve for unknowns. 2-Force Members Recognize a 2FM when analyzing a 2D equilibrium problem and use this knowledge to more quickly and easily solve the problem. FBDs & 3D rigid body equilibrium Correctly model typical 3D support reactions.

Apply force and moment equilibrium equations, in vector form, to find force and moment reactions on a 3D body or system. Friction: static/kinetic, impending motion

When friction forces are seen in equilibrium problems, correctly decide which friction relationship to use.

Friction: slip vs. tip Determine if a given problem depicts slip or tip and apply the appropriate relationships to solve. If a given problem does not specify slip or tip, analyze both situations to determine which situation will actually happen.

Frames & Machines Solve frame & machine problems by analyzing systems strategically and with internal consistency. Concepts not yet tested Statically Indeterminate Frames Recognize a statically indeterminate frame and formulate the appropriate geometric constraint equation.

Apply force-deflection, consistent w/ the FBDs, to express the geometric constraint equation in terms of forces. Trusses: Method of Joints Recognize a truss problem. Strategically apply the method of joints to solve for forces in each member. Trusses: Method of Sections Strategically apply the method of sections to solve for forces in the members you are looking for.

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EM121 Statics and Mechanics of Materials I

Final Exam – Spring 2011-2012 A Page 1 of 17

Final Exam - Spring 2011-2012

Name : ____________________________ Section Number : __________

Record all your answers to the multiple choice problems (1-15) by filling in the appropriate circle. All multiple choice answers will be graded from these markings, not the marking on the problem pages. Therefore, carefully transcribe your answers. No guessing on the multiple choice problems - you must show all your work for full credit.

Problem Answers Score

1 a b c d e f /5 2 a b c d e f /5 3 a b c d e f /5 4 a b c d e f /5 5 a b c d e f /5 6 a b c d e f /5 7 a b c d e f /5 8 a b c d e f /5 9 a b c d e f /5 10 a b c d e f /5 11 a b c d e f /5 12 a b c d e f /5 13 a b c d e f /5 14 a b c d e f /5 15 a b c d e f /5 16 /20 17 /40 18 /40 19 /40

Total /215

Show all work for credit

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EM121 Statics and Mechanics of Materials I

Final Exam – Spring 2011-2012 A Page 2 of 17

For problems 1-15, circle the best answer and then copy it to the coversheet. You must show all work for full credit. Problem 1

The projection of force vector ji ˆ2ˆ3 +− N in the direction of the position vector r shown at right is: a) 10 N b) 0.94 N c) -0.94 N d) -10 N e) None of the above

Problem 2 A rectangular plate is supported as shown at right. The tension in cable CD is expressed in vector form as

kjiTCDˆ128ˆ96ˆ120 −+−= N. The moment about point A

created by this force is: a) kji ˆ8.28ˆ0.48ˆ4.38 +− N⋅m b) 65 N⋅m c) 68 N⋅m d) kji ˆ8.28ˆ8.28ˆ68.7 ++− N⋅m e) None of the above

8 m

7 m

x

y

r

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Final Exam – Spring 2011-2012 A Page 3 of 17

Problem 3 For the system shown at right, identify the correct free body diagram for the frictionless pulley

Problem 4

We push on a cabinet of weight, W with horizontal force, P. From the analysis shown, we have found the critical height “h” that is the transition between slip and tip. (If the force is above h the cabinet will tip and below h the cabinet will slip). If we are able to lower the center of gravity of the cabinet, the critical height, h, will ΣFx=0

P-µN=0

ΣFy=0

W-N=0

ΣMcorner=0

(b/2)(W)-Ph=0

a) increase b) decrease c) stay the same d) can’t tell from given

75 lb

A

B C

D

E

a)

75

Dx

Dy

b) 75

Dx

Dy

c)

75

75

Dx

Dy

d)

75

75

Dx

Dy

M

e) none shown

f=µN

P W

N

FBD

Center of Gravity

b

h

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Final Exam – Spring 2011-2012 A Page 4 of 17

Problem 5 Consider the glued joint shown below:

What is the normal stress in the glue? (a) 1410 psi (b) 2740 psi (c) 595 psi (d) 251 psi (e) 1160 psi (f) None of the above. Problem 6

In a ship unloading operation, a 3500-lb automobile is supported by a cable. A rope is tied to the cable at A and pulled in order to center the automobile over the intended position. The angle between the cable and the vertical is 2°, while the angle between the rope and the horizontal is 30°. If the tension in the cable is 3574 pounds, what tension is needed in the rope for equilibrium? a) 125 lb b) 144 lb c) 249 lb d) 2490 lb e) None of the above

25o

1.5”

1”

5000 lb 5000 lb

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Final Exam – Spring 2011-2012 A Page 5 of 17

Problem 7 The external reaction at C of the simple cantilever truss shown at right is: a) 28 kips → b) 28 kips ← c) 36 kips → d) 36 kips ← e) None of the above

Problem 8

For the truss with external reactions shown at right, the force in member BC is a) Zero b) hwF 2 (compression) c) hwF (tension) d) hwF 2 (tension) e) None of the above

h

A

B C

D

E

F/2 F/2 F

w w

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Final Exam – Spring 2011-2012 A Page 6 of 17

Problem 9

A uniform bar, ABCD, whose cross sectional area is 0.5 in2 is loaded with the forces shown.

The tensile stress in section BC of the bar is a) 2000 psi b) 1000 psi c) 6000 psi d) 4000 psi e) 8000 psi

f) None of the above

Problem 10

In a tug of war contest between two fraternities, each of the two teams is pulling with 600 pounds of force. Assume that the cross-sectional area of the rope is 0.1 in2 and is made of a material with a failure strength of 15000 lb/in2. The factor of safety (FOS) is

a. 25 b. 5 c. 2.5 d. 1.25 e. 0.8 f. None of the above

500 lb 1000 lb 3000 lb

A B C D

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Problem 11 What is the y coordinate of the centroid of the object shown at right? a) 1.85 in b) 1.67 in c) 1.33 in d) 1.15 in e) None of the above

Problem 12

A bar (A=0.4 in2, L = 40 in, E = 10x106 psi) fits into a gap between two rigid supports because its length is 0.01 in less than the gap. The temperature is increased in a way that would cause the bar to increase in length by 0.015 in due to thermal expansion if the wall was not present.

The stress in the bar after the full temperature increase is: a) 0 psi b) 1250 psi C c) 1250 psi T d) 2500 psi C e) 2500 psi T f) None of the above

NOT TO SCALE

L = 40 in, A = 0.4 in2, E=10 x 106 psi

Initial gap = 0.01 in

x

y

3 in

3 in 3 in

2 in

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Final Exam – Spring 2011-2012 A Page 8 of 17

Problem 13 Consider the rigid beam supported by two elastic bars as shown. It pivots about a pin support on its left end.

Which statement is true regarding the forces in IDENTICAL members A and B? a) FA = FB b) FA = (0.5) FB c) FA = (2.0) FB d) FA = (0.2) FB e) None of the above

A B

10 in 20 in 20 in

P, the load

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Problem 14

A 100 lb block sits on a rough horizontal surface. A 20 lb force acts horizontally as shown. The frictional force between the block and the ground is

a) 0lb b) 10lb c) 20lb d) 30lb e) 40lb f) none of these/can’t tell from given Problem 15

We have a block on a slope that is at incipient (impending) slip. The correct equation for the friction force is

a) f=µsN b) f=µkN c) f<µsN d) f>µsN e) can’t tell from given

P

µs=0.4 µk=0.3

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Final Exam – Spring 2011-2012 A Page 10 of 17

Problem 16 (20 pts) A tension joint has the basic layout shown in the diagram below. (All dimensions on the diagram are in inches.) A single pin of diameter 0.5 inches (A = 0.2 in2) holds together the yoke and bar as shown. The material has an ultimate strength in tension of 80 ksi, and an ultimate strength in shear of 45 ksi. The load P has a value of 10000 lb. By examining the shear stress in the pin, the normal stress in the yoke, and the normal stress in the bar, find the overall factor of safety for the joint.

P=10,000 lb

P=10,000 lb

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Final Exam – Spring 2011-2012 A Page 11 of 17

Problem 16 (cont.)

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Final Exam – Spring 2011-2012 A Page 12 of 17

Problem 17 (40 pts) For the system shown at right, determine the reactions at the fixed support O (assume it is welded there).

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Final Exam – Spring 2011-2012 A Page 13 of 17

Problem 17 (cont.)

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Final Exam – Spring 2011-2012 A Page 14 of 17

Problem 18 (40 pts)

A landing gear mechanism is shown in the accompanying figure. You may assume that it is in static equilibrium in the configuration shown. You may neglect all of the weight except that of the strut and wheel (225 lb) which is located at point G. The force F, applied to member BCD, drives the retraction.

Find: a. The force in link AB. (Hint: Use a coordinate system aligned with the wheel strut.) b. The magnitude of the pin force at the pin at P. c. The force F required for static equilibrium. (Hint: Now a use a coordinate system aligned with the BC axis.)

a 2.5” b 36” c 6” d 16” e 4”

A

a

b

c

20o

B

40o

C

d

G

225 lb

70o

D F

e

P

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Final Exam – Spring 2011-2012 A Page 15 of 17

Problem 18 (cont.)

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Final Exam – Spring 2011-2012 A Page 16 of 17

Problem 19 (40 pts) Consider the A-frame shown. Note: ABC is a single member, CDE is a single member, and BDF is a single member. This is not a truss.

(a) Calculate the pin forces at B, C and D. (You can neglect the weight of the structural members.)

(b) The pins joining the A-frame are in single shear. The pins at B, C, and D have cross-sectional areas of 0.25 in2. They will fail when the shear stress reaches 20,000 psi. The other pins have much larger cross-sectional areas and will not fail first. Determine whether failure occurs, and whether the failure is at pin B, C, or D.

1.5 ft 1.5 ft 1.5 ft 1.5 ft

2.0 ft

2.0 ft

A

B

C

D

E

F

5000 lb

I am NOT a truss. I am a frame!

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Final Exam – Spring 2011-2012 A Page 17 of 17

Problem 19 (cont.)

Answers to the multiple-choice questions from the Spring 2011-2012 Final Exam.

1. c 2. d 3. c 4. c 5. c 6. b 7. a 8. b 9. e 10. c 11. d 12. b 13. d 14. c 15. a