1. No category

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.
y
x
TAB =
(2 points)
TAC =
(3 points)
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.
y
x
FB =
Increase
(3 pts)
Decrease
Remain the Same
(Circle one)
(2 pts)
NAME: ____________________________________
ME 270 – Sample Final Exam
PROBLEM 1C. (5 points)
3.8 ft
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.
y
x
T
FBC =
or
C
(Circle one)
(3 pts)
Zero-Force Members =
(2 pts)
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.
C
C
B
D
E
A
B
D
y
x
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?
μs = 0.7
P=
Increase
(3 pts)
Decrease (Circle One)
(2 pts)
ME 270 – Sample Final Exam
NAME: ____________________________________
z
PROBLEM 2. (25 points)
TBC
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.
TBD
FIND:
a) Draw a free body diagram of the mast on the artwork provided
y
below. (4 pts)
x
b) Express tension vectors TBC and TBD 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)
z
y
x
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).
C x = Fn =
(1 pt)
Cy = V =
(2 pts)
Mc =
(2 pts)
PROBLEM 3B. (5 points)
FIND: Determine the average shear stress in the
20 mm-diameter pin at A and the 30 mmdiameter pin at B that supports beam AB.
(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.
τ=
(3 pts)
ϵ=
(2 pts)
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.
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 B y reaction. (5 pts)
d) Determine the normal stress distribution just to the right of the B y 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.
( AVG )A = 34.0 MPa
( AVG )B = 17.7 MPa
3C.
 = 191 MPa
 = 9.554 x 104 mm/mm
3D.  outer
4. 
= 0.345MPa
= 960,000 (0.0025 - y2 )
 inner = 0.276MPa
 = 1,920,000 y