MSE 4220/6220 – Mechanical Behavior of Materials
Homework 01 - SOLUTION
Please contact instructor (vblouin@clemson.edu) if you identify a typo in this document or if you have any
question.
Problem 1 (25 points)
(1) Give the definition of “Force” and the corresponding SI unit and English unit.
A force is an action applied at a point, on a line, on a surface or on a body of a structure in a given
direction. It is measured in Newtons (N) or pounds (lbs).
(2) What is the difference between “Normal Force” and “Shear Force”? Give an example for each.
A normal force is applied on a surface in a direction perpendicular (normal) to the surface.
Example: The reaction force from the ground applied under the sole of your shoe due to your
weight is a normal force.
A shear force is applied on a surface in a direction parallel to the surface.
Example: The friction force from the ground applied under the sole of your shoe is a shear force.
(3) What is the difference between “External Force” and “Internal Force”? Give an example for each.
External forces are applied on the “system of interest” by external effects that are not part of the
system. Examples: gravity, supports, structural and non-structural elements connected to or in
contact with the system.
Internal forces are applied on a part of the system by the other part. The two parts of the system
are separated by an imaginary cut. Example: Consider a bar simply supported by simple supports at
both ends. The bar bends by its own weight. Consider an imaginary cut through the middle of the
bar that separates the bar into two parts A and B. The forces applied on part A by part B are the
internal forces.
(4) Give the definition of “Stress” and the corresponding SI unit and English unit.
A stress is a force per unit surface area. It is measured in N/m2 or Pascals (Pa) for SI and pounds per
square inch (psi) for English units.
(5) What is the difference between “Normal Stress” and “Shear Stress”? Give an example for each.
Normal stress is the normal force per unit surface area (normal to the surface of interest).
Example: The normal stress applied under my shoes (10 in2 per shoe) under my own weight is:
170 lbs / (2*10 in2) = 8.5 psi
The shear stress is the shear force per unit surface area (parallel to the surface).
Example: Assume that I am brave enough to stand straight up on the roof of my house that has a
pitch of 45 degrees. Assume that the friction coefficient between the sole of my shoes and the roof
shingles is 0.5. If my shoes start sliding, the shear force is equal to the normal force multiplied by
the friction coefficient. That is shear force = 170*cos(45)*0.5 = 60.1 lbs
Then the shear stress = 60.1/(2*10 in2) = 3.05 psi.
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Problem 2 (40 points)
A vertical post is composed of two parts as shown.
The top part has a cross-sectional area of 30 mm2
and the bottom part has a cross-sectional area of
50 mm2. The post is fixed at the bottom (point E), a
mass is attached at points A and C. B and D are the
mid-points of each rod. You may neglect the mass
of the rod.
(1) Determine all external forces applied on
the rod and draw a free-body diagram of
the rod.
A
430 kg
4.22
kN
B
B
C
There are 3 external forces:
The 430 kg mass creates a downward force at A of
430*9.81 = 4.22 kN.
The 150 kg mass creates a downward force at C of
150*9.81 = 1.47 kN.
The reaction force at E is also in the y-direction and
can be determined using the static equilibrium
equation. Sum of forces in the positive y-direction
must be equal to zero:
RE – 4.22 – 1.47 = 0 → RE = 5.69 kN
A
150 kg
C
D
D
E
E
1.47
kN
RE
y
x
(2) Determine the internal forces applied on the cross-section through point B. Make sure to use the
step-by-step procedure studied in class (even if it is trivial).
Step 1 - The system is selected (that is the two bars and masses)
Step 2 - The coordinate system is already defined.
A
4.22
Step 3 - The free-body diagram (FBD) of the whole system was
kN
drawn in the previous question.
Step 4 - The external forces were calculated in the previous
B
question.
N
Step 5 - The point of interest (B) was selected for us.
Step 6 - Cut through B.
Step 7 - Select a side: top
Step 8 - Draw FBD of the selected part with internal force. Only N is needed since the problem is
one-dimensional. The shear force V and bending moment M are zero.
Step 9 - Use static equilibrium equation to determine the internal force N. Sum of forces in positive
y-direction equals zero:
-4.22 – N = 0 → N = -4.22 kN
The negative sign makes sense since we expect the normal force N at B to counteract the
downward weight at A. So the normal force is expected to be positive upward.
(3) Determine the average normal stress applied on the cross-section through B.
 = N/A = -4,220/30 = -140.67 N/mm2 = -140.67 MPa
The negative sign makes sense since we expect the stress to be compressive.
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(4) Repeat (2) and (3) for D.
Since this is a relatively simple problem, we can skip steps 1 to 5 of
the step-by-step procedure.
A
Step 6 - Cut thru D
Step 7 - Select a side: top
Step 8 - Draw the FBD
Step 9 - Use static equilibrium equation to determine the internal
force N
-4.22 – 1.47 – N = 0 → N = -5.69 kN
4.22
kN
B
1.47
kN
C
 = N/A = -5,690/50 = -113.8 MPa (Compressive)
D
N
Note 1: It would have been slightly easier to select the bottom part.
The result should be the same.
Note 2: When calculating the stress , we used the minus sign in the formula. However, it would
have been acceptable to drop the minus sign as long as we specified the fact that the stress is
“compressive” at D using intuition and engineering understanding of the problem. In general, you
must pay strict attention to signs when calculating forces and moment but not when calculating
stresses.
Note 3: This is another reason for dropping the sign when calculating the stress: If we had selected
N positive upward at D (since it is an arbitrary decision), we would have ended up with N = +5,69
kN instead of -5.69 kN. The resulting stress would have been +113.8 MPa instead of -113.8 MPa
and we would have specified “Compressive” since N is positive and pointing towards the crosssection.
(5) Describe the stress through B as precisely as possible using some of the following terms: normal,
shear, bending, longitudinal, torsional, transversal, compressive, and tensile.
The stress through B is a normal compressive stress applied in the longitudinal direction or a
longitudinal normal compressive stress.
(6) Repeat (2) and (3) but this time, use the other side of the cut
through B and show that you obtain the same results. That is if you
selected the bottom part of the cut, use the top and vice versa.
Step 7 - Use the bottom side
Step 8 - Draw the FBD. It is customary to use N pointing outward
from the cross-section. This N is the same as in question (2)
because of “action and reaction”.
Step 9 - Static equilibrium using the sum of forces with positive
upward orientation:
N – 1.47 + 5.69 = 0
→ N = -4.22 kN
Stress:
3
N
B
C
1.47
kN
D
E
5.69
kN
 = -4,220/30 = -140.67 MPa (Compressive)
The results of N and  are the same as in questions (2) and (3).
Problem 3 (10 points)
Repeat question (3) of Problem 2 but without neglecting the weight of the rod. Assume that the rod is
made of aluminum and the length of each part is 2 meters.
Assume that the density of aluminum is 2,700 kg/m3.
We need to recalculate the normal force at B. The half rod above B creates a downward gravity force equal
to: (2,700kg.m3)*(1m)*(40e-6m2)*(9.81m/s2) = 10.59 N
The normal force N at B is then:
-4,220.00 – 10.59 – N = 0 → N = -4,230.59 N
The stress is then:
 = N/A = -4,230.59/30 = -141.02 MPa
This result is very similar to that of (2), which confirms that gravity is negligible.
Note: In question (1) of Problem 2, the force at A was simplified to 4.22 kN (i.e., 3 significant digits). The
result of Problem 3 would be more accurate if you used 5 or 6 significant digits: 4.2183 kN
Problem 4 (25 points)
A uniform cable is attached at both ends with a mass hanging at the middle as shown.
(1) What are the four pieces of information that you need to determine the normal stress in the cable?
The four parameters needed to determine the normal stress in the cable are:
Set 1:
M (hanging mass)
L (length of cable)
D (distance between the two walls)
A (cross-sectional area of the cable)
Actually, other sets of parameters may be sufficient. These parameters can be calculated from the four
parameters M, L, D, A of Set 1:
Set 2: Only 3 parameters are needed.
M (hanging mass)
 (angle of the cable)
A (cross-sectional area of the cable)
Set 3: Only 3 parameters are needed.
Rx (reaction in x at each wall)
Ry (reaction in y at each wall)
A (cross-sectional area of the cable)
Set 4: Only 3 parameters are needed.
Ry (reaction in y at each wall)
 (angle of the cable)
A (cross-sectional area of the cable)
Set 5: Only 2 parameters are needed.
T (tension of the cable)
A (cross-sectional area of the cable)
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(2) Assume numerical values for each and calculate the corresponding stress in the cable. List all
values clearly.
Depending on which set of parameters was used:
R
Set 1:
M = 10 kg → Mg = 98.1 N

L=1m
D = 0.8 m
A = 0.0001 m2
The reaction force at the wall is in the direction of the
cable (since a cable is not able to sustain bending). Knowing L and D, the angle of the cable can be
calculated: cos() = D/L →  = 36.9 deg
The vertical component Ry of the reaction force is half of M: Ry = Mg/2 = 49.0 N
Knowing the angle, the reaction force can be calculated: R = Ry/sin() = 81.7 N
Since R is applied in the direction of the cable, the internal force is equal to R → N = 81.7 N
The stress is then:  = N/A = 81.7/0.0001 = 0.82 MPa
Set 2:
M = 10 kg → Mg = 98.1 N
 = 40 deg
A = 0.0001 m2
The vertical component Ry of the reaction force is half of M: Ry = Mg/2 = 49.0 N
Knowing the angle, the reaction force can be calculated: R = Ry/sin() = 76.2 N
Since R is applied in the direction of the cable, the internal force is equal to R → N = 76.2 N
The stress is then:  = N/A = 76.2/0.0001 = 0.76 MPa
Set 3:
Rx = 60 N
Ry = 50 N
A = 0.0001 m2
The reaction at the wall can be calculated from the components: R = sqrt(Rx2+Ry2) = 78.1 N
Since R is applied in the direction of the cable, the internal force is equal to R → N = 78.1 N
The stress is then:  = N/A = 78.1/0.0001 = 0.78 MPa
Set 4:
Ry = 50 N
 = 40 deg
A = 0.0001 m2
Knowing the angle, the reaction force can be calculated: R = Ry/sin() = 77.8 N
Since R is applied in the direction of the cable, the internal force is equal to R → N = 77.8 N
The stress is then:  = N/A = 77.8/0.0001 = 0.78 MPa
Set 5:
T = 80 N
A = 0.0001 m2
The internal force N is also the tension of the cable: → N = 80 N
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The stress is then:  = N/A = 80/0.0001 = 0.80 MPa
(3) Does any shear stress exist in the cable? Explain why.
A cable only works in tension. It cannot resist shear and bending because of its flexibility. This
means that if shear and/or bending is applied on any part of the cable, the cable reorients itself to
remain in tension and cancel the shear force and/or the bending moment.
Problem 5 - Extra credit for undergraduate students (5 points) and required for graduate students (20
points)
A electrical pole is held in place transversally using shrouds (i.e.,
vertical cables under tension) and spreaders (i.e., horizontal rods
under compression used to push the shrouds away from the
mast). The pole has a pin connection to the ground. Assume that
the tension of the vertical part of the shrouds is 7,000 N, the
cross-section the pole is circular solid of diameter 30 cm and
uniform along the length of the pole. The cross-sectional area of
the shrouds and spreaders are 80 mm2, and 200 mm2,
respectively. Assume pin connections everywhere. Neglect
gravity.
pole
6m
spreader
6m
shroud
2m
(1) Are the forces in the pole, spreaders, and shrouds
compressive or tensile?
Usually, the shrouds are under tension to make sure that the pole does now move.
Because of the geometry of the system, the shrouds pull the pole down and put it under compression.
The shrouds are pushed away from each other by the spreaders. Therefore, the spreaders must be under
compression.
(2) Is the force in the upper part of the pole
equal to that of the lower part?
R1
The FBD’s of the three subsystems are
drawn. The angle of the top part of the
shrouds is . Since the three subsystems are
connected with each other using pin
connections, the reactions on the spreaders
are in the longitudinal direction of the
spreaders (i.e., no shear, no moment). This
means that the spreaders do not apply any
vertical force on the pole, only transversal
forces R2. Since R2 is transversal, it does not
have any effect on the longitudinal
R1

R2
7.0 kN
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R2
R2
R2
R3
R1
R2
compressive force that is applied on the pole. Therefore, the compressive force is the same in the
upper and lower parts of the pole.
(1) Calculate the stresses in the pole, spreaders, and shrouds.
First, we can calculate  given the dimensions of the system: tg() = 2/6 →  = 18.4 deg
The three subsystems are in static equilibrium. The shroud provides 2 static equilibrium equations:
Sum of vertical forces = 0 → -7.0kN + R1cos() = 0 → R1 = 7.0kN/cos(18.4) = 7.38 kN
Sum of horizontal forces = 0 → R2 – R1sin() = 0 → R2 = 7.38kN*sin(18.4) = 2.33 kN
The static equilibrium of the pole provides 2 equations but only the sum of the vertical forces is
needed to determine R3 → R3 – 2*R1*cos() = 0 → R3 = 2*7.38*cos(18.4) = 14.0 kN
This result (R3 = 14.0 kN) was obvious because the two shrouds pull the pole down, each with 7 KN.
Now the stresses can be calculated:
Bottom part of the shrouds: shroud_bottom = 7,000/80 = 87.5 N/mm2 = 87.5 MPa
Top part of the shrouds: shroud_top = 7,380/80 = 92.3 MPa
Spreaders: spreaders = 2,330/200 = 11.7 MPa
Pole: pole = 14,000/(*0.32/4) = 198.06 MPa
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