Shear and moment diagram

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1. Shear and Moment Diagrams
Ambrose - Chapter 6
MD. INTISHAR RAHMAN

2. Beam Shear
•Vertical shear:
tendency for one part
of a beam to move
vertically with respect
to an adjacent part
–Section 3.4 (Ambrose)
–Figure 3.3 =>

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. 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. Beam Shear – Example 1 (pg. 64)
•Simple beam
–Span = 20 feet
–2 concentrated loads
•Construct shear
diagram

6. Beam Shear – Example 1 (pg. 64)
1)Determine the reactions
Solving equation (3):
Solving equation (2):
Figure 6.7a =>
()( )
()( )
()( )å
å
å
´-´+´==¿+
+--==­+
=­+
)'20()'161200()'68000(03
1200800002
01
2
..
1
2
..
1
RM
RRF
F
lblb
lblb
y
x
()­==
+=
-
--
.
.
..
2
....
2
360,3
20
200,67
200,19000,48'20
lb
ft
ftlb
ftlbftlb
R
R
()­=
-+=
.
1
...
1
840,5
360,3200,1000,8
lb
lblblb
R
R

7. Beam Shear – Example 1 (pg. 64)
•Determine the shear at
various points along the beam
.
)18(
.
)8(
.
)1(
360,3200,1000,8480,5
160,2000,8480,5
480,50480,5
lb
x
lb
x
lb
x
V
V
V
-=--=
-=-=
+=-=
=
=
=

8. Beam Shear – Example 1 (pg. 64)
•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. Beam Shear – Example 2 (pg. 66)
•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. Beam Shear – Example 2 (pg. 66)
•Determine the reactions
Solving equation (3):
Solving equation (2):
()( )
()( )
()( )å
å
å
´-´+´==¿+
+-´-==­+
=­+
)'24()'166000()]'6)('12000,1[(03
000,6)'12000,1(02
01
2
..
1
2
..
1
RM
RRF
F
lbftlb
lbftlb
y
x
()­==
+=
-
--
.
.
..
2
....
2
000,7
24
000,168
000,96000,72'24
lb
ft
ftlb
ftlbftlb
R
R
()­=
-+=
.
1
...
1
000,11
000,7000,6000,12
lb
lblblb
R
R

11. Beam Shear – Example 2 (pg. 66)
•Determine the shear at
various points along the beam
( )[ ]
( )[ ]
.
)24(
.
)16(
.
)16(
.
)12(
.
)2(
.
)1(
000,7000,6000,112000,11
000,7000,6000,112000,11
000,1)000,112(000,11
000,1)000,112(000,11
000,9)000,12(000,11
000,10)000,11(000,11
lb
x
lb
x
lb
x
lb
x
lb
x
lb
x
V
V
V
V
V
V
-=+´-=
-=+´-=
-=´-=
-=´-=
=´-=
=´-=
=
+=
-=
=
=
=

12. Beam Shear – Example 2 (pg. 66)
•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 R
1
–at this location, V = 0
feetx
xV
11
)000,1(000,110
=
´-==

13. Beam Shear – Example 3 (pg. 68)
•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

14. Beam Shear – Example 3 (pg. 68)
•Determine the reactions
Solving equation (3):
Solving equation (4):
()( )
()( )
()( ) ( )( )[ ]
()( ) ( )( )[ ] )'20(4
2
32
)'32500()'28000,2()'6000,12()'4000,4(04
)'20(8
2
32
)'32500()'24000,4()'14000,12()'8000,2(03
)'32500(000,4000,12000,202
01
1
.....
2
2
.....
1
2
...
1
.
RM
RM
RRF
F
ftlblblblb
ftlblblblb
ftlblblblb
y
x
+-´-´-´-´==¿+
--´+´+´+´-==¿+
+´---+-==­+
=­+
å
å
å
å
()­==
+++-=
-
----
.
.
..
2
........
2
800,18
20
000,376
000,128000,96000,168000,16'20
lb
ft
ftlb
ftlbftlbftlbftlb
R
R
()­==
+++-=
-
----
.
..
1
........
1
200,15
'20
000,304
000,192000,56000,72000,16'20
lb
ftlb
ftlbftlbftlbftlb
R
R

15. Beam Shear – Example 3 (pg. 68)
•Determine the shear at
various points along the beam

16. Beam Shear – Example 3 (pg. 68)
•Conclusions
–max. vertical shear =
12,800 lb.
•disregard +/- notations
–shear passes through zero
at three points
•R
1
, R
2
, and under the
12,000lb. load

17. 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)

18. Bending Moment

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

20. Bending Moment – Example 1
1)Compute moments at
critical locations
–under 8,000 lb. load &
1,200 lb. load
( )
( )
....
)'16(
...
)'6(
440,13)'10000,8()'16840,5(
040,350)'6840,5(
ftlblblb
x
ftlblb
x
M
M
-
=
-
=
=´-´=¿+
=-´=¿+

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

22. Bending Moment – Example 2
1)Compute moments at
critical locations
–When x = 11 ft. and under
6,000 lb. load
( ) ( )( )[ ]
( ) ( )( )( )[ ]
...
)'16(
...
)'11(
000,564
2
12
12000,1)'16000,11(
500,60
2
1111000,1)'11000,11(
ftlbftlblb
x
ftlbftlblb
x
M
M
-
=
-
=
=+´-´=¿+
=´-´=¿+

23. 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)

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

25. 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

26. Negative Bending Moment -
Example
()( )
()( )
()( ) ( )[ ]å
å
å
´-´==¿+
+´-==­+
=­+
)'20(
2
26
)'26600(03
)'26600(02
01
2
.
1
2
.
1
RM
RRF
F
ftlb
ftlb
y
x
1)Determine the
reactions
- Solving equation (3):
- Solving equation (4):
()­==
=
-
-
.
.
..
2
..
2
140,10
20
800,202
800,202'20
lb
ft
ftlb
ftlb
R
R
()­=
-=
.
1
..
1
460,5
140,10600,15
lb
lblb
R
R

27. 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)60020(140,10460,5
540,6)60020(460,5
540)60010(460,5
860,4)6001(460,5
lb
x
lb
x
lb
x
lb
x
V
V
V
V
=´-+=
-=´-=
-=´-=
=´-=
+=
-=
=
=

28. 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.
feetx
xV
1.9
)600(460,50
=
´-==

29. 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
'20600)'20460,5(
843,24
2
1.9
'1.9600)'1.9460,5(
ftlbftlblb
x
ftlbftlblb
x
M
M
-
=
-
=
-=´-´=
=´-´=

30. 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.
( )()[ ]
feetfeetx
x
xxxxM
x
xM
x
xxM
ftlb
2.18;0
600
460,55460
)300(2
)0)(300(4)460,5(460,5
460,5300300460,50
2
600460,50
2
600)460,5(0
2
22
2
=
-
±-
=
-
--±-
=
+-=-==
÷
÷
ø
ö
ç
ç
è
æ
-==
´-´==

31. 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

32. Rules of Thumb/Review
•moment is dependent upon the shear diagram
–the area under the shear diagram = change in the
moment (i.e. A
shear 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

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