ME 323 – Mechanics of Materials
Examination #2
November 5th, 2020
Name (Print) ______________________________________________
(Last)
(First)
PROBLEM # 1 (25 points)
A simply supported beam AH is subject to a constant distributed load q over the section BC, a moment
M0 and a concentrated force P at D. The cross section of the beam is shown below.
The parameters are following: L=8 ft., q=10 lb/ft, M0=40 lb×ft, P=10 lb, b= 2 in.
a) Draw the shear force and bending moment diagrams. Mark the values at the cross sections A, B,
C, D, and H, and the maximum and minimum values along the beam.
b) Determine the stress state at the points M and N which are located at the cross section C. Sketch
their stress state on the given stress elements.
𝑦
𝑞
𝑦
𝑃
A
𝑥
𝐿/4
D
C
B
𝐿/4
H
𝐿/4
𝑀0
𝐿/4
𝑏
M
0.2 𝑏
𝑧
0.2 𝑏
N
𝑉 𝑥
𝑏
𝐂𝐫𝐨𝐬𝐬 𝐬𝐞𝐜𝐭𝐢𝐨𝐧
0
𝑀 𝑥
0
𝑦
𝑦
𝑥
𝑥
𝑥
𝑥
𝑧
𝑧
Stress element M
Stress element N
ME 323 – Mechanics of Materials
Examination #2
November 5th, 2020
Name (Print) ______________________________________________
(Last)
(First)
PROBLEM # 1
X
Iti
i
ME 323 – Mechanics of Materials
Examination #1
November 5th, 2020
Name (Print) ______________________________________________
(Last)
(First)
PROBLEM # 2 (25 points).
Y
A
MA
B
X
L
The linearly elastic beam shown in the figure supports a couple MA at end A. The beam is homogeneous,
with Young’s modulus E, and has constant cross-section, with moment of inertia I.
(a) Using the following free body diagram, write the equations of equilibrium and identify whether
the structure is statically determinate or indeterminate.
Y
MA
A
B
MB
X
RA
L
RB
Using the second-order integration method:
(b) Determine the bending moment M(x) of the beam (as a function of the reactions at A, the external
loads and the geometric parameters).
(c) Determine the slope v’(x) and deflection v(x) of the beam.
(d) Indicate the boundary conditions at supports A and B.
(e) Solve for the reaction at A, i.e., RA.
0
gMB Ito EFy
tYfda
E
4
3 unknowns AIRBNB
staticallyIndeterminate
EIN aMatRaX
second order
Msxt'zRaEtQ
EIN
EINE Has
Boundary
Koko
Conditions
Nero
N'Chao
Solvefor Ra
Matkal MB
MexrMatRAX
five
integration method
0
MB
RB
2equations
FRA
f
ff
Md
RatRB
Qz
Ma
tQetQz
tRa
o
traffic Lao
ftp.LtlzRXLTQ
MAITRA
9
HisLt'zRaL2tCc
MAtztf R.ae
0
2 2
0
o
system
of equations
Rae 32M
ME 323 Examination # 2 AM
SOLUTION
PROBLEM #4 (25 Points)
Part A – 5 points
The beam shown below has a second area moment I for its cross-section, and is made of a material
with a Young’s modulus of E. Using the method of superposition:
a) Determine the reaction on the beam at location C. Leave your answer in terms of the load P.
b) Determine the deflection of the beam at end D. Leave your answer in terms of P, a and EI.
Superposition tables are provided.
y
P
C
B
D
x
a
y
C
B
a
y
Cy
P
B
C
D
D
x
a
a
x
a
loading 1
a
loading 2
Using the displacement at C using the loadings shown above:
v a = v1 a + v2 a
()
()
()
1⎧ 2
⎫C y 1 ⎧ 2
⎫P
0 = ⎨ a ⎡⎣3 a − a ⎤⎦ ⎬ + ⎨ a ⎡⎣3 2a − a ⎤⎦ ⎬
6⎩
⎭ EI 6 ⎩
⎭ EI
⎡2
5 ⎤ a3
5
= ⎢ C y + P⎥
⇒ Cy =− P
6 ⎦ EI
2
⎣6
The displacement at D using the loadings shown above:
v a = v1 a + v2 a
() ()
()
()
() ( )
()
2
1⎧ 2
⎫C y 1 ⎧
⎫P
0 = ⎨ a ⎡⎣3 2a − a ⎤⎦ ⎬ + ⎨ 2a ⎡⎣3 2a − 2a ⎤⎦ ⎬
6⎩
⎭ EI 6 ⎩
⎭ EI
3
3
3
3
⎡ 5 ⎛ 5 ⎞ 16 ⎤ Pa
5a
16 a
7 Pa
=
Cy +
P = ⎢ ⎜− ⎟ + ⎥
=
6 EI
6 EI
⎣ 6 ⎝ 2 ⎠ 6 ⎦ EI 12 EI
() ( )
( ) ( ) ( )
Page 1 of 4
ME 323 Examination # 2 AM
SOLUTION
PROBLEM #4 (continued)
Part B – 8 points
The loading on the beam shown below is not provided in the figure. The shear force V(x) for the beam is given
below the beam, with the shear force being provided in terms of kips and the position variable x in ft. In
addition, it is known that the bending moment at the left end is M(0) = 0, and there are no concentrated couples
applied to the beam at any locations except, possibly, at H.
For the shear force diagram provided:
a) Draw the bending diagram M(x) on the axes provided.
b) Show the loading on the beam in the figure below.
No justification is needed for your answers.
40
20
40
20
80 / 3
B
C
D
H
V ( x)
40
20
0
−20
x
−40
M ( x)
60
x
0
−30
0
1.5
3
6
9
Page 2 of 4
ME 323 Examination # 2 AM
SOLUTION
PROBLEM #4 (continued)
Part C – 6 points
The bending moment diagram for a loaded beam is shown below. The beam is known to have the triangular
cross section shown below. Provide a justification for each answer.
a) At what location(s) on the beam does the maximum tensile normal stress exist? Provide both x and Y
components of the location of this point(s). You are not asked to solve for this value of stress.
b) At what location(s) on the beam does the maximum compressive normal stress exist? Provide both x and
Y components of the location of this point(s). You are not asked to solve for this value of stress.
M ( x)
Y
40
20
0
−20
x
−40
a
Z
A
B
D
M ( x)
C
beam cross-section
H
Y
40
At location B along beam: M = 40
•
At Y = 0: σ =
•
At Y = a: σ =
(
20
0
−20
) = 40 a (tensile )
M −a /3
I
(
a
−40
3 I
) = 80 a (compressive )
M 2a /3
I
x
3 I
Z
beam cross-section
At location C along beam: M = 20
•
At Y = 0: σ =
•
At Y = a: σ =
(
) = 20 a (compressive )
M −a /3
I
(
3 I
) = 40 a (tensile )
M 2a /3
I
3 I
Page 3 of 4
ME 323 Examination # 2 AM
SOLUTION
PROBLEM #4 (continued)
Part D – 6 points
The cross-sections for Beams 1 and 2 are shown below. Let I1 and I2 represent the centroidal second area
moment (about the z-axis) for beams 1 and 2, respectively. Each beam is experiencing the same shear force of V
at the cross section. Let τ 1B and τ 2B be the shear stress at points B on Beams 1 and 2, respectively.
a) Circle the correct answer below in regard to the relative sizes of I1 and I2 . You are not asked to
provide numerical values for these second area moments, or justification for your answers.
•
I1 > I2
•
I1 = I2
•
I1 < I2
The vertical sections (the “webs”) of the two cross-sections have the same second area
moments. Likewise, the horizontal sections (the “flanges”) have the same second area
moments. Therefore, the second area moments are the same.
b) Circle the correct answer below in regard to the relative sizes of τ 1B and τ 2B . You are not asked
to provide numerical values for these stresses, or justification for your answers.
•
τ 1B > τ 2B
•
τ 1B = τ 2B
•
τ 1B < τ 2B
The first area moments (Q) for the webs are the same, as well as for the flanges.
The “thickness” of the cross sections are the same (2b). Therefore, the shear
stresses are the same.
y
y
R
z
neutral axis
R
h
B
3b
z
neutral axis
b
b
RR
Beam 1
h
B
b
RR
Beam 2
Page 4 of 4