NAME: ____________________________________

ME 270 – Sample Final Exam

PROBLEM 1 (25 points) – Prob. 1 questions are all or nothing.

PROBLEM 1A. (5 points)

FIND: A 2000 N crate (D) is suspended using ropes AB and AC and is in

static equilibrium. If θ = 53.13˚, determine the tension in ropes AB

and AC.

PROBLEM 1B. (5 points)

FIND: Plate AB is supported by a pin support at A and

a rocker support at B and is in static equilibrium.

For the loading shown (neglect the weight of the

plate), determine the magnitude of the reaction

force at B. If the 100 in-lb couple were shifted to

point A, would FB increase, decrease or remain the same.

TAB = (2 points)

TAC =

(3 points)

FB = (3 pts)

Increase Decrease Remain the Same (Circle one) (2 pts)

x

y

x

y

NAME: ____________________________________

ME 270 – Sample Final Exam

PROBLEM 1C. (5 points)

FIND: Truss ABCDEFG is loaded with a single 600 lb force at

joint C and is in static equilibrium. Determine the

magnitude of the force in member BC and whether it is

in tension or compression. List all zero-force members.

PROBLEM 1D. (5 points)

FIND: The A-Frame shown is supported by a rocker support at

A and a pin support at E and is in static equilibrium. On the

schematics below, complete a free-body diagram of each

member of the A-Frame.

FBC = T or C (Circle one) (3 pts)

Zero-Force Members = (2 pts)

x

y

B D

A

B

C

E

D

C

x

y

3.8 ft

NAME: ____________________________________

PROBLEM 1E. (5 points)

FIND: For the 50 lb chest shown, determine the force P needed to tip the chest, assuming the

dimension d = 3 ft. If the 30˚ angle were increased, would it increase or decrease the probability of

tipping the chest?

P = (3 pts)

Increase Decrease (Circle One) (2 pts)

μs = 0.7

NAME: ____________________________________

ME 270 – Sample Final Exam

PROBLEM 2. (25 points)

GIVEN: The vertical mast supports the 4-kN force and is

constrained by two cables BC and BD and by a

ball-and-socket connection at A.

FIND:

a) Draw a free body diagram of the mast on the artwork provided

below. (4 pts)

b) Express tension vectors BCT and BDT in terms of their

unknown magnitudes and their known unit vectors. (4 pts)

c) Determine the tension magnitudes TBC and TBD. (8 pts)

d) Determine the magnitudes of the reactions at the ball-and-socket A. (9 pts)

BCT BDT

x

z

y

y

x

z

NAME: ____________________________________

PROBLEM 3 (25 points) – Prob. 1 questions are all or nothing.

PROBLEM 3A. (5 points)

FIND: Determine the internal forces on cross-section C.

of the beam shown (i.e., the axial and shear-force

and the bending-moment).

PROBLEM 3B. (5 points)

FIND: Determine the average shear stress in the

20 mm-diameter pin at A and the 30 mm-

diameter pin at B that supports beam AB.

Cx = Fn = (1 pt)

Cy = V = (2 pts)

Mc = (2 pts)

(τAvg)A = (3 pts)

(τAvg)B = (2 pts)

NAME: ____________________________________

PROBLEM 3C. (5 points)

FIND: Beam AB rests on the two short posts (AC and BD).

Post AC has a diameter of 20 mm and is made of

steel. Determine the stress and strain in post AC if

it is made of steel and has a diameter of 20 mm.

Assume Esteel = 200 GPA.

PROBLEM 3D. (5 points)

FIND: Tube AB has an inner diameter or 80 mm and an

outer diameter of 100 mm. Given the torque

applied in the diagram, determine the shear stress

acting on the inner and outer walls of the tube.

τ = (3 pts)

ϵ = (2 pts)

τouter = (3 pts)

τinner = (2 pts)

NAME: ____________________________________

PROBLEM 4 (25 points)

GIVEN: Beam AB is loaded as shown. Assume the

beam has a rectangular cross-section of

50 mm wide and 100 mm high.

FIND:

a) Determine the reaction forces at supports

A and B. (4 pts)

b) Sketch the shear-force and bending-moment diagrams. (6 pts)

c) Determine the shear stress distribution just to the right of the By reaction. (5 pts)

d) Determine the normal stress distribution just to the right of the By reaction. (5 pts)

NAME: ____________________________________

V (kN)

X (m)

V (kN)

X (m)

Sample Final Exam Answers

1A. TAB = 1500 N TAC = 2500 N

1B. FB = 15.3 lbs Remain the same

1C. FBC = 632 lbs Zero-Force Members: BF, DE, CD, EF, CE

1D. FBD

1E. P = 26.8 lbs Increase

2. TBC = 4.47 kN TBD = 4.90 kN

A = -2 i + 0j +8k kN

3A. CX = -300 lbs Cy = -10.42 lb Mc = 718.75 lb-in

3B. A( ) = 34.0 MPaAVG ( ) = 17.7 MPaAVG B

3C. = 191 MPa 4 = 9.554 x 10 mm/mm

3D. = 0.345MPaouter = 0.276MPainner

4. 2 = 960,000 (0.0025 - y ) = 1,920,000 y