1. 1
BEAMS
SHEAR AND MOMENT

2. 2
Beam Shear
 Shear and Moment
Diagrams
 Vertical shear:
tendency for one part
of a beam to move
vertically with respect
to an adjacent part

3. 3
Beam Shear
 Magnitude (V) = sum of vertical forces on
either side of the section
– can be determined at any section along the
length of the beam
 Upward forces (reactions) = positive
 Downward forces (loads) = negative
 Vertical Shear = reactions – loads
(to the left of the section)

4. 4
Beam Shear
 Why?
 necessary to know the maximum value of
the shear
– necessary to locate where the shear changes
from positive to negative
• where the shear passes through zero
 Use of shear diagrams give a graphical
representation of vertical shear throughout
the length of a beam

5. 5
Beam Shear –
 Simple beam
– Span = 20 feet
– 2 concentrated loads
 Construct shear
diagram

6. 6
Beam Shear – Example 1
1) Determine the reactions
Solving equation (3):
Solving equation (2):
Figure 6.7a =>
  
  
  





















)
'
20
(
)
'
16
1200
(
)
'
6
8000
(
0
3
1200
8000
0
2
0
1
2
.
.
1
2
.
.
1
R
M
R
R
F
F
lb
lb
lb
lb
y
x
 








.
.
.
.
2
.
.
.
.
2
360
,
3
20
200
,
67
200
,
19
000
,
48
'
20
lb
ft
ft
lb
ft
lb
ft
lb
R
R
 





.
1
.
.
.
1
840
,
5
360
,
3
200
,
1
000
,
8
lb
lb
lb
lb
R
R

7. 7
Beam Shear – Example 1 (pg. 64)
 Determine the shear at various
points along the beam
.
)
18
(
.
)
8
(
.
)
1
(
360
,
3
200
,
1
000
,
8
480
,
5
160
,
2
000
,
8
480
,
5
480
,
5
0
480
,
5
lb
x
lb
x
lb
x
V
V
V

















8. 8
Beam Shear – Example 1
 Conclusions
– max. vertical shear = 5,840 lb.
• max. vertical shear occurs at greater
reaction and equals the greater reaction
(for simple spans)
– shear changes sign under 8,000 lb. load
• where max. bending occurs

9. 9
Beam Shear – Example 2
 Simple beam
– Span = 20 feet
– 1 concentrated load
– 1 uniformly distr. load
 Construct shear diagram,
designate maximum shear,
locate where shear passes
through zero

10. 10
Beam Shear – Example 2
 Determine the reactions
Solving equation (3):
Solving equation (2):
  
  
  






















)
'
24
(
)
'
16
6000
(
)]
'
6
)(
'
12
000
,
1
[(
0
3
000
,
6
)
'
12
000
,
1
(
0
2
0
1
2
.
.
1
2
.
.
1
R
M
R
R
F
F
lb
ft
lb
lb
ft
lb
y
x
 








.
.
.
.
2
.
.
.
.
2
000
,
7
24
000
,
168
000
,
96
000
,
72
'
24
lb
ft
ft
lb
ft
lb
ft
lb
R
R
 





.
1
.
.
.
1
000
,
11
000
,
7
000
,
6
000
,
12
lb
lb
lb
lb
R
R

11. 11
Shear and Moment Diagrams

12. 12
Beam Shear – Example 2
 Determine the shear at various
points along the beam
 
 
 
  .
)
24
(
.
)
16
(
.
)
16
(
.
)
12
(
.
)
2
(
.
)
1
(
000
,
7
000
,
6
000
,
1
12
000
,
11
000
,
7
000
,
6
000
,
1
12
000
,
11
000
,
1
)
000
,
1
12
(
000
,
11
000
,
1
)
000
,
1
12
(
000
,
11
000
,
9
)
000
,
1
2
(
000
,
11
000
,
10
)
000
,
1
1
(
000
,
11
lb
x
lb
x
lb
x
lb
x
lb
x
lb
x
V
V
V
V
V
V







































13. 13
Beam Shear – Example 2
 Conclusions
– max. vertical shear = 11,000 lb.
• at left reaction
– shear passes through zero at some
point between the left end and the
end of the distributed load
• x = exact location from R1
– at this location, V = 0
feet
x
x
V
11
)
000
,
1
(
000
,
11
0






14. 14
Beam Shear – Example 3
 Simple beam with
overhanging ends
– Span = 32 feet
– 3 concentrated loads
– 1 uniformly distr. load
acting over the entire beam
 Construct shear diagram,
designate maximum shear,
locate where shear passes
through zero

15. 15
Beam Shear – Example 3
  
  
    
 
 
    
 
  )
'
20
(
4
2
32
)
'
32
500
(
)
'
28
000
,
2
(
)
'
6
000
,
12
(
)
'
4
000
,
4
(
0
4
)
'
20
(
8
2
32
)
'
32
500
(
)
'
24
000
,
4
(
)
'
14
000
,
12
(
)
'
8
000
,
2
(
0
3
)
'
32
500
(
000
,
4
000
,
12
000
,
2
0
2
0
1
1
.
.
.
.
.
2
2
.
.
.
.
.
1
2
.
.
.
1
.
R
M
R
M
R
R
F
F
ft
lb
lb
lb
lb
ft
lb
lb
lb
lb
ft
lb
lb
lb
lb
y
x













































 













.
.
.
.
2
.
.
.
.
.
.
.
.
2
800
,
18
20
000
,
376
000
,
128
000
,
96
000
,
168
000
,
16
'
20
lb
ft
ft
lb
ft
lb
ft
lb
ft
lb
ft
lb
R
R
 













.
.
.
1
.
.
.
.
.
.
.
.
1
200
,
15
'
20
000
,
304
000
,
192
000
,
56
000
,
72
000
,
16
'
20
lb
ft
lb
ft
lb
ft
lb
ft
lb
ft
lb
R
R

16. 16
Determine the reactions
Solving equation (3):
Solving equation (4):

17. 17
Beam Shear – Example 3
 Determine the shear at various
points along the beam
 
 
 
 
 
 
  .
.
.
.
.
.
.
)
32
(
.
.
.
.
.
.
.
)
28
(
.
.
.
.
.
.
)
28
(
.
.
.
.
.
.
)
22
(
.
.
.
.
.
)
22
(
.
.
.
.
.
)
8
(
.
.
.
.
)
8
(
000
,
4
'
32
500
000
,
12
000
,
2
800
,
18
200
,
15
000
,
6
'
28
500
000
,
12
000
,
2
800
,
18
200
,
15
800
,
12
'
28
500
000
,
12
000
,
2
200
,
15
800
,
9
'
22
500
000
,
12
000
,
2
200
,
15
200
,
2
'
22
500
000
,
2
200
,
15
200
,
9
'
8
500
000
,
2
200
,
15
000
,
6
'
8
500
000
,
2
0
lb
ft
lb
lb
lb
lb
lb
x
lb
ft
lb
lb
lb
lb
lb
x
lb
ft
lb
lb
lb
lb
x
lb
ft
lb
lb
lb
lb
x
lb
ft
lb
lb
lb
x
lb
ft
lb
lb
lb
x
lb
ft
lb
lb
x
V
V
V
V
V
V
V



























































18. 18
Beam Shear – Example 3
 Conclusions
– max. vertical shear = 12,800
lb.
• disregard +/- notations
– shear passes through zero at
three points
• R1, R2, and under the 12,000lb.
load

19. 19
Bending Moment
 Bending moment: tendency of a beam to bend due
to forces acting on it
 Magnitude (M) = sum of moments of forces on
either side of the section
– can be determined at any section along the length of the
beam
 Bending Moment = moments of reactions –
moments of loads
• (to the left of the section)

20. 20
Bending Moment

21. 21
Bending Moment – Example 1
 Simple beam
– span = 20 feet
– 2 concentrated loads
– shear diagram from
earlier
 Construct moment
diagram

22. 22
Bending Moment – Example 1
1) Compute moments at
critical locations
– under 8,000 lb. load &
1,200 lb. load
 
  .
.
.
.
)
'
16
(
.
.
.
)
'
6
(
440
,
13
)
'
10
000
,
8
(
)
'
16
840
,
5
(
040
,
35
0
)
'
6
840
,
5
(
ft
lb
lb
lb
x
ft
lb
lb
x
M
M


















23. 23
Bending Moment – Example 2
 Simple beam
– Span = 20 feet
– 1 concentrated load
– 1 uniformly distr. Load
– Shear diagram
 Construct moment
diagram

24. 24
Bending Moment – Example 2
1) Compute moments at
critical locations
– When x = 11 ft. and under
6,000 lb. load
    
 
    
 
  .
.
.
)
'
16
(
.
.
.
)
'
11
(
000
,
56
4
2
12
12
000
,
1
)
'
16
000
,
11
(
500
,
60
2
11
11
000
,
1
)
'
11
000
,
11
(
ft
lb
ft
lb
lb
x
ft
lb
ft
lb
lb
x
M
M




















25. 25
Negative Bending Moment
 Previously, simple beams subjected to positive
bending moments only
– moment diagrams on one side of the base line
• concave upward (compression on top)
 Overhanging ends create negative moments
• concave downward (compression on bottom)

26. 26
Negative Bending Moment
 deflected shape has
inflection point
– bending moment = 0
 See example

27. 27
Negative Bending Moment -
Example
– Simple beam with
overhanging end on right
side
• Span = 20’
• Overhang = 6’
• Uniformly distributed load
acting over entire span
– Construct the shear and
moment diagram
– Figure 6.12

28. 28
Negative Bending Moment -
Example
  
  
    
 




















)
'
20
(
2
26
)
'
26
600
(
0
3
)
'
26
600
(
0
2
0
1
2
.
1
2
.
1
R
M
R
R
F
F
ft
lb
ft
lb
y
x
1) Determine the
reactions
- Solving equation (3):
- Solving equation (4):
 






.
.
.
.
2
.
.
2
140
,
10
20
800
,
202
800
,
202
'
20
lb
ft
ft
lb
ft
lb
R
R
 




.
1
.
.
1
460
,
5
140
,
10
600
,
15
lb
lb
lb
R
R

29. 29
Negative Bending Moment -
Example
2) Determine the shear at
various points along
the beam and draw
the shear diagram
.
)
20
(
.
)
20
(
.
)
10
(
.
)
1
(
600
,
3
)
600
20
(
140
,
10
460
,
5
540
,
6
)
600
20
(
460
,
5
540
)
600
10
(
460
,
5
860
,
4
)
600
1
(
460
,
5
lb
x
lb
x
lb
x
lb
x
V
V
V
V


























30. 30
Negative Bending Moment -
Example
3) Determine where the
shear is at a
maximum and where
it crosses zero
– max shear occurs at the right
reaction = 6,540 lb.
feet
x
x
V
1
.
9
)
600
(
460
,
5
0






31. 31
Negative Bending Moment -
Example
4) Determine the
moments that the
critical shear points
found in step 3) and
draw the moment
diagram
  
 
  
  .
.
.
)
20
(
.
.
.
)
1
.
9
(
800
,
10
2
20
'
20
600
)
'
20
460
,
5
(
843
,
24
2
1
.
9
'
1
.
9
600
)
'
1
.
9
460
,
5
(
ft
lb
ft
lb
lb
x
ft
lb
ft
lb
lb
x
M
M
















32. 32
Negative Bending Moment -
Example
4) Find the location of the inflection
point (zero moment) and max.
bending moment
 since x cannot =0, then we use x=18.2’
 Max. bending moment =24,843 lb.-ft.
  
 
feet
feet
x
x
x
x
x
x
M
x
x
M
x
x
x
M ft
lb
2
.
18
;
0
600
460
,
5
5460
)
300
(
2
)
0
)(
300
(
4
)
460
,
5
(
460
,
5
460
,
5
300
300
460
,
5
0
2
600
460
,
5
0
2
600
)
460
,
5
(
0
2
2
2
2


































33. 33
Rules of Thumb/Review
 shear is dependent on the loads and reactions
– when a reaction occurs; the shear “jumps” up by the
amount of the reaction
– when a load occurs; the shear “jumps” down by the
amount of the load
 point loads create straight lines on shear diagrams
 uniformly distributed loads create sloping lines of
shear diagrams

34. 34
Rules of Thumb/Review
 moment is dependent upon the shear diagram
– the area under the shear diagram = change in the
moment (i.e. Ashear diagram = ΔM)
 straight lines on shear diagrams create sloping
lines on moment diagrams
 sloping lines on shear diagrams create curves on
moment diagrams
 positive shear = increasing slope
 negative shear = decreasing slope

35. 35
Typical Loadings
 In beam design, only
need to know:
– reactions
– max. shear
– max. bending moment
– max. deflection