Quiz 12
Question 1 What is most nearly the moment of inertia of the beam shown below about its centroid in
the y direction.
(A) 3.66 in^4
(B) 20.5 in^4
(C) 133.89 in^4
(D) 154.39 in^4
This problem is solved by obtaining the moment of inertia of the 3 component rectangles and then
using the parallel axis theorem to transfer those moments of inertia to the common axis of the overall structure.
1- Find the moment of inertia of each component area
Finding the moment of intertia of each componant about its axis
Part
b
h I
1
3 1
2.25
2
0.75 7
0.246094
3
6 1
18
2- Find the centroid of the composite structure. This is done by taking the area weighted average
of the centroids of each component
Finding each composit centroid as area weighted average of componants
Part
A
centroid
A*cent
1
3
8.5
25.5
2
5.25
4.5
23.625
3
6
0.5
3
sums
14.25
52.125
A*cent/A
over-all centroid
3.657895 y
3- Now use the parallel axis theorem to transfer the individual moments of inertia to the common
axis.
Apply parallel axis theorem to shift axis of each componant inertia
Part Comp C Overall C
Diff
A
Diff^2*A
I own C
1
8.5 3.657895 4.842105
3 70.33795
2.25
2
4.5 3.657895 0.842105 5.25 3.722992 0.246094
3
0.5 3.657895
-3.15789
6
59.8338
18
133.8947 sum
The answer 154.39 corresponds to D.
Total I
72.58795
3.969085
77.8338
154.3908
Question 2 If a load is distributed as shown below on a 50 ft beam that has the cross-section shown in
question 1, which of the 4 diagrams below most nearly represents the moment diagram?
(A)
(B)
(C)
(D)
B
This problem does not require us to construct a moment diagram but only to identify which one it is.
We note first that because the ends are pin connected we must have zero moment at the ends. We also
know that a moment will have units of ft*lbs. Only B and D meet these conditions.
We next note that just the first 1000 lb load at 10 feet from point A will have to generate a moment
10,000 ft*lbs. Only B has numbers big enough to come from loads this size. The small moments in D
look more like they must come from a unit weight load – typical of an influence line. The numbers on D
are much too small. It has to be B.
(Of course you could construct the reactions at A and B and then the shear diagram which could be
integrated to produce the moment diagram).
Question 3 One of the diagrams above on question 2 represent the moment influence line for the point
position of maximum stress on the beam. If a 1000 lb load moves across the beam what will most nearly
be the maximum bending moment seen in the beam?
(A)
(B)
(C)
(D)
6.25 ft*lbs
32,500 ft*lbs
32,506 ft*lbs
38,750 ft*lbs
We already know that (B) from question 2 above is the moment diagram for the static load on the beam
so we know that the maximum bending stress is at mid span (surprise – surprise) and is 32,500. We also
know that we have a moving load of 1,000 lbs that will generate its own moment.
Several solutions are possible
1- Method 1 – Observation – the moment produced by a 1000 lb load is not 0. Since we were at
32,500 ft*lbs without the added load only (C) and (D) are possible answers. There is no way a
1,000 lb could produce only 6 ft*lbs of moment so (C) is not the answer. We pick (D) by
elimination.
2- Method 2 – Recognize the influence line. We suspected (D) on question 2 above was the
influence line. If we convince ourselves with (D) is the moment influence line at C for a moving
unit load we can do just like is routinely done in bridge design. Take the peak moment at C for
the influence line 6.25 ft*lbs for a unit weight of 1, multiply it by 1,000 (1,000 lbs is 1,000*1) and
add the influence line result to the moment diagram. 32,500 + 6,250 = 38,750 which is (D).
3- Method 3 – Construct our own influence line at C by computing the reactions at A and B and
then using those reactions*12.5 to compute the influence line moment diagram
X
ra
0
5
Rb
1
0.8
sum a C
0
0.2
MB
0
2.5
10
0.6
0.4
5
12.5
0.5
0.5
6.25
With the moment influence line in hand we now solve just like in method 2.
4- Method 4 – Do static check with the load at one point. A load of 1,000 lbs will reach its greatest
lever arm in the middle of the beam. Using simple statics we know we get 500 lbs of support
from each support and each support has a lever arm of 12.5. 500*12.5 = 6,250. Add 6,250 to
32,500 = 38,750 which corresponds to D.