ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
Please review the following statement:
I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam.
Signature: ______________________________________
INSTRUCTIONS
Begin each problem in the space provided on the examination sheets. If additional space is required,
use the white lined paper provided to you.
Work on one side of each sheet only, with only one problem on a sheet.
Each problem is worth 25 points.
Please remember that for you to obtain maximum credit for a problem, it must be clearly presented,
i.e.






The only authorized exam calculator is the TI-30IIS
The allowable exam time for Final Exam is 120 minutes.
The coordinate system must be clearly identified.
Where appropriate, free body diagrams must be drawn. These should be drawn separately
from the given figures.
Units must be clearly stated as part of the answer.
You must carefully delineate vector and scalar quantities.
If the solution does not follow a logical thought process, it will be assumed in error.
When handing in the test, please make sure that all sheets are in the correct sequential order
and make sure that your name is at the top of every page that you wish to have graded.
Instructor’s Name and Section:
Sections:
J Jones 9:50-10:50AM
J Jones Distance Learning
Problem 1 __________
Problem 2 __________
Problem 3 __________
Problem 4 ___________
Total _______________
ME 270 Final Exam – Summer 2014
Page 1
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
PROBLEM 1 (25 points) – Prob. 1 questions are all or nothing.
1A. Sphere E has a weight of 100N and is supported by cable
CBA and spring CD. Determine the magnitudes of the
tension in cable CB (assume the pulley is ideal) and the
tension in spring CD. (5 pts)
T CB =
(3pts)
TCD =
(2pts)
1B. Member AB has a 900N force and 500 N-m couple loading
the member. It is held in static equilibrium by a collar at A
and a roller support at B. The collar at A can freely slide in
the y-direction but motion in the x-direction and any rotation
about point A is resisted by the collar. Determine the
magnitudes of the reactions at collar A and roller B. (5 pts)
Ax =
(1pt)
By =
(1pt)
MA =
(3pts)
ME 270 Final Exam – Summer 2014
Page 2
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
1C. For the truss shown, identify all zero-force members
and calculate the magnitudes of the loads in members
CD and AE and specifiy whether each is in tension or
compression. (6 pts)
Zero-Force Members =
(2pts)
FCD =
T or C (Circle One) (2pts)
FAE =
T or C (Circle One) (2pts)
1D. Frame ABC is designed as a compound beam which has a pin connection between members AB
and BC. The beam is supported by a fixed support at A and a rocker support at C. For the loads
shown, determine the forces acting at B and C in vector form on member BC. (Hint: Sketch the
individual free body diagrams first.) (4 pts)
BonBC =
(2pts)
ConBC =
(2pts)
ME 270 Final Exam – Summer 2014
Page 3
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
1E. Determine the total angle of wrap around pegs A, B, and C collectively and calculate the smallest
force P required to lift the 100 N crate. Assume the coefficient of friction between the cable and each
peg is μ s = 0.1. (5 pts)
β=
(3pts)
Pmin =
(2pts)
ME 270 Final Exam – Summer 2014
Page 4
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
PROBLEM 2. (25 points)
GIVEN: The sign has a weight of 1000N with a center of mass
at G. The sign is held in static equilibrium by a ball
and socket support at A and cables BC and BD.
D
D
FIND:
a) On the sketch provided, complete the free body diagram
of the sign. (3 pts)
D
b) Write vector expressions for the forces in cables BC and BD in terms of their unknown magnitudes
and their known unit vectors. (4 pts)
TBC =
(2pts)
T BD =
(2pts)
ME 270 Final Exam – Summer 2014
Page 5
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
c) Determine the magnitudes of the tensions in cables BC and BD. (10 pts)
TBC =
(5pts)
TBD =
(5pts)
d) Determine the magnitudes of the reactions at the ball and socket support. (8 pts)
Ax =
(2pts)
Ay =
(3pts)
Az =
(3pts)
ME 270 Final Exam – Summer 2014
Page 6
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
PROBLEM 3. (25 points)
3A. A wood specimen is subjected to an average normal stress of 2 ksi during a tensile test.
Determine the axial load P applied to the specimen. Also, find the average shear stress
developed along section a-a of the specimen. (5pts)
P=
 τa-a avg
(2pts)
=
(3pts)
3B. A static analysis of beam AB results in FBC = 12.5kN and A = -7.5 i + 20 j kN . Determine the
average shear stress (in kPa) in the 20 mm-diameter pin at A and the 30 mm-diameter pin at B
that support the beam. (Hint: Note the type of connection used on each end of the beam.) (5 pts)
 τA avg =
 τB avg =
ME 270 Final Exam – Summer 2014
(3pts)
(2pts)
Page 7
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
3C. Determine the magnitude of the torque (in kN-m) pulleys A and B
create in shaft AB and calculate the maximum shear stress (in kPa)
due to this torsional load. Assume the shaft has a diameter of 40 mm.
(5 pts)
T=
(2pts)
τ Max =
(3pts)
3D. The cross-section of a beam is “T-shaped” with the dimensions shown. The location of the
centroid is identified on the artwork. Determine the second-area moment of inertia about the
x-axis (Ix). Based on the shape of the cross-section, which of these statements are true
y
(Ix > Iy , Ix < Iy, Ix = Iy). No calculations are necessary. (5 pts)
x
Ix =
Ix > Iy ;
(3pts)
Ix < I y ; Ix = Iy
ME 270 Final Exam – Summer 2014
(Circle One) (2pts)
Page 8
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
3E. For the shaded area bounded by the y-axis, y = 1 and x = y2; determine the area and the
second moment area of inertia about the y-axis (Ix). Hint:Use a horizontal differential element.
(5 pts)
x = y2
A=
(2pts)
Ix =
(3pts)
ME 270 Final Exam – Summer 2014
Page 9
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
PROBLEM 4. (25 points)
Given: Beam ABCDE is loaded as shown and is held in static equilibrium by a pin support at A and a
roller support at D. The beam cross-section is “T-shaped” and has the dimensions shown and a
second moment of inertia of I x = 13.8 x 10-6 m 4 . (Note – NA refers to the neutral axis.)
FIND:
a) Sketch a free-body diagram of the beam and determine the magnitudes of the reactions at A and
D. (5 pts)
4 kN
2 kN/m
21 kN-m
NA
3m
A
B
3m
1m
1m
C
E
D
FBD (1 pt)
3m
A
B
1m
1m
C
D
3m
E
Ay =
(2pts)
Dy =
(2pts)
ME 270 Final Exam – Summer 2014
Page 10
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
4b) On the axes provided, sketch the shear-force and bending-moment diagrams of the beam.
(10pts)
4 kN
2 kN/m
21 kN-m
3m
A
1m
B
3m
1m
C
D
E
V(kN)
x(m)
M(kN-m)
x (m)
ME 270 Final Exam – Summer 2014
Page 11
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
4c) In which segment(s) of the beam does pure bending occur (2 pts)
AB
BC
CD
DE
None
(Circle all that apply)
4d) In the segment of the beam where pure bending exists, determine the magnitudes of the
maximum tensile bending stress and the maximum compressive bending stress. (8 pts)
 σmax T
 σmax C
=
(4pts)
=
(4pts)
ME 270 Final Exam – Summer 2014
Page 12
ME 270 – Summer 2014 Final Exam
NAME (Last, First): ________________________________
Summer 2014 Final Exam – Equation Sheet
Normal Stress and Strain
σx =
x
Fn
A
V(x) = V(0) + ∫ p(ϵ)dϵ
0
−My
σx (y) =
I
σx ∆L
εx =
=
E
L
εy = εz = − ϑεx
−y
εx (y) =
ρ
σfail
FS =
σallow
x
M(x) = M(0) + ∫ V(ϵ)dϵ
0
Buoyancy
FB  gV
Fluid Statics
p  gh
Feq  pavg  Lw 
Shear Stress and Strain
τ(ρ) =
Belt Friction
TL
 e
TS
V
A
τ=
Shear Force and Bending Moment
Tρ
J
Distributed Loads
L
Feq   w  x  dx
τ = Gγ
0
G=
E
2(1 + ϑ)
xFeq   x w  x  dx
γ=
δs π
= −θ
Ls 2
Centroids
L
0
Second Area Moment
I = ∫ y 2 dA
c
ci
y
y A
y
A
i
ci
i
A
1
I=
bh3
12
π
I = r4
4
 x dA
 dA
x A
x
A
x
IB = IO + AdOB 2
Polar Area Moment
π 4
r
Circle
2
π
J = (ro4 − ri4 ) Tube
2
J=
i
x V
In 3D, x 
V
i
i
i
i
Centers of Mass
 x dA
 dA
x  A
x
 A
cm
x
cmi
i
i
i
i
ME 270 Final Exam – Summer 2014
i
i
i
ci
Circle
c
i
i
Rectangle
 y dA
 dA
i
 y dA
 dA
y  A
y
 A
cm
y
i
cmi
i
i
i
i
i
i
Page 13
ME 270 – Spring 2014 Exam 1
NAME (Last, First): ________________________________
ME 270 Final Exam Solutions – Summer 2014
1A.
TCB = 115 N
TCD = 57.5 N
1B.
Ax = 0 N
By = 900 N
MA = - 1486 N-m
1C. Zero-Force Members: AF, EF, BC
FCD = 400 lbs
C
FAE = 667 lbs
1D.
BonBC = 4 j kN
ConBC = 4 j kN
1E.
β = 360o = 2 π rad.
Pmin = 187 N
C
2A. FBD
2B.
TBC = TBC (0.333 i - 0.667 j + 0.667k)
TBD = TBD (- 0.707 i - 0.707 j)
2C.
TBC = 750 N
TBD = 354 N
2D.
Ax = 0 N
A y = 750 N
3A.
P = 4000 lbs = 4 kips
 τa-a avg
Az = 500 N
= 250 psi
3B.  τ B avg = 17,684 kPa
3C.
T = 0.60 kN-m
τmax = 47,770 kPa
3D.
I x = 33.3 in 4
Ix > Iy
3E.
A = 1/3m2 = 0.333 m2
I y = 1/5m4 = 0.20 m4
4A. FBD
A y = 4 kN
D y = 6 kN
4B. Shear-force and bending-moment diagrams
4C. BC, CD
4D.  τmax T = 136,522 kPa
[Type text]
 τmax C
= 102,391 kPa
Page 14