Shear & Moment Diagrams Examples

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CE 331, Fall 2007
Shear & Moment Diagrams Examples
1 / 7

We will analyze bending in a beam due to two types of loads: dead loads and live loads.
Dead loads typically represent the weight of the structure itself. Live loads typically
represent the weight of the structure's contents (people in an office building, merchandise in a
warehouse, trucks on a bridge). Live loads differ from dead loads in that the location of live
loads is uncertain. This will be discussed further in Step 3 of the example problem below.
Example Problem 1

Given the loads and beam configuration shown below, calculate the maximum moment due
to all loads.
• uniform distributed dead load (w
D
) = 0.50
klf
applied to entire beam
• uniform distributed live load (w
L
) = 1.00
klf
applied along either Span 1 only, Span 2
only, or Spans 1 and 2






The most accurate solution would be to calculate the bending moment for the following three
cases:

Case I: DL + LL on Span 1 only



Case II: DL + LL on Span 2 only



Case II: DL + LL on Spans 1 & 2


Structural engineers use computer programs to do just this. For this hand-calculation example, we
will calculate a bending moment which should be close to the actual maximum moment due to dead
plus live loads. We will use the following procedure:
1. Draw the beam to scale (horizontally). The resulting shear and moment diagrams will also be
drawn to this horizontal scale. These scaled drawings can be used to check our calculations
for reasonableness.
17 ft
8 ft
Span 1
Span 2
w
D
= 0.5
klf

w
L
= 1.0
klf

w
D
= 0.5
klf

w
L
= 1.0
klf

w
D
= 0.5
klf

w
L
= 1.0
klf

CE 331, Fall 2007
Shear & Moment Diagrams Examples
2 / 7

2. Draw the shear and moment diagram due to dead load. Note the magnitude and location of
the maximum bending moment, M
D
.

3. Calculate the moment due to live load, M
L
.
We will assume that the maximum moment due
to dead plus live loads (M
D+L
)
occurs at the location of the maximum moment due to dead
load (M
D
)
. This assumption is often correct, and even when not, yields a maximum total
moment (M
D+L
)
that is reasonably close to the actual total moment.
4. Calculate the total moment due to dead + live load, M
D+L


Solution.

1. Determine the horizontal scale for your sketch (use maximum):
• 7” across engineering paper, 5 little squares per inch, = 35 little squares
• 25 foot-long-beam / 35 little squares = 0.71 feet per little square
• Round 0.71 up to nearest of: 1, 2, 5, 10, 20, 50, … Use 1 little square per foot

( or 1” = 5’)
See sketch on next page.
2. Calculate the max moment due to dead load (max M
D
)
2.1 Calculate reactions
ChecksF
RRM
RRM
ftklfkk
v
kftftftklf
kft
ft
ftklf
,0)25)(5.0(31.319.9,0
31.3,0)17)(()5.4)(25)(5.0(,0
19.9,0)17)(()
2
25
)(25)(5.0(,0
supportleftsupportleft
trsupportright
supportrightsupportright supportleft
=−+=Σ
==−=Σ
==−=Σ










2.2 Draw the shear diagram to scale (see sketch below).
2.3. Draw the moment diagram to scale (see below). Note: Change in moment = area
under shear diagram.

Resultant =
(
0.5
klf
)(
25
ft
)
= 12.5
k
12.5
12.5
4.5 ft
CE 331, Fall 2007
Shear & Moment Diagrams Examples
3 / 7



max M
D
= 16.0
k-ft
at Support 2


3. Calculate the max. moment due to live load (M
L
) at the location of the max. moment due
to dead load (M
D
).
3.1 Determine where to place the live load to cause the max M
L
at the middle of Span 1.
As mentioned on Page 1, the location of live loads is variable. Although it's
physically possible that live loads are located on only part of a span, structural
W = 0.5 klf
17 ft
8 ft
9.19
k
3.31
k

V, k

3.31

-5.19
4.0
ftk
k
/
5.0
31.3
=
10.38
ft
6.62
ft
ftkkft −
== 96.10)31.3)(62.6(
2
1
26.94
k
-
ft
16.0
k
-
ft
M, k-ft

10.96
k
-
ft
-16.0
k
-
ft
V = 0
k
at beam end,
which it should
M = 0
k
at beam end,
which it should
CE 331, Fall 2007
Shear & Moment Diagrams Examples
4 / 7

engineers typically apply distributed live loads to the entire span. The choices are
therefore to apply live load to:


Span 1 only,


Span 2 only, or


Spans 1 and 2.
To determine which of the above span loading patterns will cause the max M
L
at
Support 2, we look at the deflected shape of the beam that will cause the max.
bending moment at this location (see below).
Bend beam over Support 2 to cause negative bending (since M
D
= -16.0
k-ft
)






Load spans to cause the deflected shape above.





3.2 Calculate the moment at Support 2 due to live load (M
L
)
We could calculate the reactions, draw the V and then the M diagram;
OR use one of the formulas below:
Simply-supported Beam with Uniform Load:







w
w L
2
/ 8
M
L
w
L
= 1.0
klf

CE 331, Fall 2007
Shear & Moment Diagrams Examples
5 / 7

Cantilever Beam with Uniform Load:








Therefore, M
L
= w
L
L
2
/ 2 = (1.0
klf
)(8
ft
)
2
/ 2 = 32.0
k-ft

M
L
= -32.0
k-ft


4. Calculate the total moment at Support 2, M
D+L
.
M
D+L
= M
D
+ M
L
= -16.0
k-ft
+ -32.0
k-ft
= -48.0
k-ft

M
D+L
= -48.0
-ft



Check: Envelope of maximum moments due to M
D
and M
L
from computer-aided analysis:

Max. M = -48
k-ft
at Support 2 (same as hand-calcs.)

w
w L
2
/ 2
M
L
CE 331, Fall 2007
Shear & Moment Diagrams Examples
6 / 7

Example Problem 2.
(same as Problem 1 except overhang = 7
ft
)


uniform distributed dead load (w
D
) = 0.50
klf
applied to entire beam


uniform distributed live load (w
L
) = 1.00
klf
applied along either Span 1 only, Span 2
only, or Spans 1 and 2




M
D
:
max M
D
= 12.46
k-ft
in "middle" of Span 1

17 ft
7 ft
Span 1
Span 2
W = 0.5 klf
17 ft
7 ft
8.47
k
3.53
k

V
, k

3.53

-4.97
3.50
9.94
ft
7.06
ft
24.70
k
-
ft
12.5
k
-
ft
M, k-ft

12.46

-12.24
12.46
k
-
ft

OKF
RM
RM
ftklfkk
v
k
k
A
,0)24)(5.0(57.853.3,0
53.3,0
47.8,0
11
2
=−+=Σ
=→=Σ
=→=Σ

=0
k
,
O
K

=0
k-ft
,
O
K

CE 331, Fall 2007
Shear & Moment Diagrams Examples
7 / 7

M
L
:
Bend beam at middle of Span 1 to cause positive bending (since M
D
= +12.46
k-ft
)





Live load Span 1 to cause the deflected shape above.






Therefore, M
L
= w
L
L
2
/ 8 = (1.0
klf
)(17
ft
)
2
/ 8 = 36.1
k-ft

M
L
= +36.1
k-ft

M
D+L
:
M
D+L
= M
D
+ M
L
= 12.46
k-ft
+ 36.1
k-ft

M
D+L
= 48.6
k-ft



Check:
Envelope of maximum moments due to M
D
and M
L
from computer-aided analysis:
From the computer-aided analysis:


the max M
D
= 12.46
k-ft
at 7.06
ft
from the left support


the max M
L
= 36.1
k-ft
at 8.50
ft
from the left support


the max M
D+L
= 48.1
k-ft
at 7.97
ft
from the left support
w
L
= 1.0
klf