-
Shear Forces and Bending Moments
Planar (2-D) Structures:All loads act in the same plane and all
deflections occurs in the same plane (x-y plane)
Associated with the shear forces and bending moments are normal
stresses and shear stresses.
-
Types of Beams, Loads, and Supports
-
Procedure for determining shear force and bending moment
Determine the reactions using the equilibrium conditions of the
overall structure
Cut the beam at the cross section at which shear force and
bending moment are to be determined. Draw a free-body diagram
Set up equilibrium equations of the F.B.D. to determine shear
force and bending moment at the cross section
Draw the shear force and bending moment diagrams
-
Sign conventions for shear force V and bending moment M.
Sign Convention
-
Find the shear force and the bending moment at cross sections to
the left and to the right of the midpoint.
Reactions
( )
LMPR
LMPR
MLPLRM
PRRF
OA
OB
OBA
BA
=+=
=
==
=+==
43
4
04
0
00
To the left of the midpoint:
28
042
0
4 00
2
O
AL
OA
MPLM
MLPLRM
LMPVVPRF
=
=+
+
==
====
-
To the right of the midpoint:
28
042
0
4 00
2
O
OAL
OA
MPLM
MMLPLRM
LMPVVPRF
+=
=+
+
==
====
The shear force does not change (because the vertical force
acting on the free body do not change) but the bending moment
increases by an amount equal to MO.
To the left of the load P
V
M
416
04
0
4 00
4
O
AL
OA
MPLM
MLRM
LMPVVRF
=
=+
==
====
-
LMP O
43
LMP O
4
416OMPL
28OMPL
28OMPL +
-
Find the shear force and bending moment at a distance x from the
free end of the beam.
L32
Reactions at B (concentrate the distributed load)
( )
6
03
22
0
2
02
0
2LqM
MLLqLRM
LqRLqRF
OB
BO
BA
OB
OB
=
=+
==
====
2LqO
At a distance x
Lxqq O=
Intensity of the distributed load.
-
V(x)
2LqO
M(x)
6
2LqO
-
Find the shear force and bending moment at a distance x=15ft
from A.
Reactions (concentrate the distributed load)
( )( )
( ) ( )( ) ( )( )( )lbRlbR
ftftftlbftlbftRMlbRR
ftftftlblbRRF
AB
BA
BA
BA
11000 900001530/200914000240
200000624/200140000
==
===
=+
=++==
xVVxF
200110000200110000
=
===
Shear Forces and bending moments:From A to the concentrated load
P: 0
-
( ) ( )
xxM
xxxMM
11000100
110002
2000
2 +=
+==
x
x
From the concentrated load P to B: 9
-
Relationships between Loads, Shear Stresses and Bending
Moments
Distributed Loads
( ) 00 =+== dVVqdxVFVertical qdxdV
=
Bending Moment (ccw as positive)
( ) ( ) 02
0 =+++
== dMMdxdVVdxqdxMM V
dxdM
=
( )PV
VVPVFVertical=
=+==1
1 00
The concentrated loads cause abrupt changes in the shear force
wherever they are located.
( ) ( )
dxVVdxdxPM
MMdxVVdxPMM
11
11
2
02
0
++
=
=+++
==
As the differentials are small, the bending moment does not
change as we pass through the point of application of a
concentrated load.
-
Loads in the form of couples:
( )0
00
1
1
=
=+==V
VVVFVertical
The shear force does not change at the point of application of a
couple.
( ) ( )O
O
MMMMdxVVMMM
=
=++++==1
11 00
The bending moment changes abruptly at the point of application
of a couple.
-
Revisit the following example : Find the shear force and bending
moment at a distance x from the free end of the beam.
Lxqq O=
-
LxqV
CVthenxFor
CLxqVdx
LxqdV
Lxqq
dxdV
Lxqq
o
oO
OO
2
00____0__2
2
1)0(
1
2
=
===
+==
===
LxqM
CMthenxFor
CLxqM
LxqV
dxdM
o
oOx
6
00__0_62
3
2)0(
2
32
=
===
+===
-
Shear Force and Bending Moments Diagrams
Axial Forces
Torsion
-
Shear Force and Bending Moment Diagrams
Finding the reactions:
-
Reactions
Shear-force and bending moment diagrams for a simple beam with
several concentrated loads.
-
Simple beam with a uniform load over part of the span.
-
The maximum bending moments occurs at the location where the
shear force is zero
-
Cantilever beam with two concentrated loads.
-
Cantilever beam with a uniform load.
qxVCqxqdxV =+== 1( )
2
22
2
2
xqM
CxqM
dxqxVdxM
=
+=
==
For x=0 then V(x=0)=0
For x=0 then M(x=0) =0
For x=L then
Vdx
dMqdxdV
==
2
2qLM =
-
2
1
qLqxV
CqxqdxV
qdxdV
+=
+==
=
xqLxqM
CxqLxqM
dxqLqxVdxM
Vdx
dM
22
22
2
2
22
+=
++=
+==
=
For x=0 then V(x=0) =RA
For x=0 then M(x=0) =0
For x=L/2 then M=qL2/8
-
CqdxV
qdxdV
==
==
0
22112122113
32121332
111
112211221
1111
0 0
aPaPxPxPxRMaPaPDDxPxPxRMPPRCaxa
aPxPxRMaPDDxPxRMPRCaxa
xRMDDxRMRCax
Ax
AxA
Ax
AxA
Ax
AxA
++=+=+==
+==+==
==+==
( )DCxM
dxCVdxM
Vdx
dM
+=
==
=
321443
21332
1221
11
0
PPPRCaxaPPRCaxa
PRCaxaRCax
A
A
A
A
==
==
-
( )( )LPPPRxPxPxPxRM
LPPPRDMLxDxPxPxPxRMPPPRCaxa
AAx
AL
Ax
A
321321
3214
4321
321443
0
====
+==
-
Beam with an overhang.
Reactions
( )( )( )( )( ) ( )
=
=
=+=
=+=
kRkR
RM
RRF
C
B
BC
CBVertical
25.125.5
00.12161840.1
040.1
-
Reactions( )( )( )
( )( )( ) ( )( ) ( )( )( ) ( )
=
=
=+=
=+=
kNR
kNR
RM
mmkNRRF
B
A
AB
BAVertical
5.12
5.7
08120.51102720.5
02/0.52
@
Simple beam
( ) ( ) 5.75 5 1 +=+== xVCxqdxV xx
Vdx
dMqdxdV
==
For 0
-
Method of Superposition
This can be treated as the superposition of four simple
cases.
Thetotalshearforcediagramisthesumoftheindividualshear
forcediagramsforeachcase.
Thetotalbendingmomentdiagramisthesumoftheindividualbendingmomentdiagramsforeachcase.
-
( )655.2865.2 645.24255.720
===
=
=
xVxVxVx
xVx
V
x
x
x
x
x
( )( )( )( )( )( )
( )( )( ) ( )( ) ( )
0.805.275.22
6651255.786
0.105.21255.7 640.105.21255.742
5.25.720
2
2
+=
=
+==+==
=
=
xxM
xxxxMx
xxxMxxxxMx
xxMx
M
x
x
x
x
x
x
0860 64042020
====
=
x
x
x
x
x
VxVxVxVx
V
208620 64042020
====
=
x
x
x
x
x
MxMxMxMx
M
Reactions=
=
kNRkNR
B
A
5.125.7
-
( )655.2865.2 645.24255.720
===
=
=
xVxVxVx
xVx
V
x
x
x
x
x
( )( )( )( )( )( )
( )( )( ) ( )( ) ( )
0.805.275.22
6651255.786
0.105.21255.7 640.105.21255.742
5.25.720
2
2
+=
=
+==+==
=
=
xxM
xxxxMx
xxxMxxxxMx
xxMx
M
x
x
x
x
x
x
0860 64042020
====
=
x
x
x
x
x
VxVxVxVx
V
208620 64042020
====
=
x
x
x
x
x
MxMxMxMx
M
0.605.275.2860.305.2 640.105.242
5.25.720
2
2
+=+=+=
=
=
xxMxxMxxMx
xxMx
M
x
x
x
x
x
( )655.2865.2 645.24255.720
===
=
=
xVxVxVx
xVx
V
x
x
x
x
x
-
=====
=kNRVxkNRVx
VxV
Bx
Bx
x
x
25.5201225.5124
040
( )( )
====
==
2125.5420122125.54124
040
xxRMxxxRMx
MxM
Bx
Bx
x
x
Reactions
=
=
kR
kR
C
B
25.1
25.5
( )( )
====
===
44201244124
40
qVxqVx
xqxVxV
x
x
x
x
( )( )( )( )
+==+==
==
=842420128424124
2240
22
xxqMxxxqMx
xqxMx
M
x
x
x
x
===
=020120124040
x
x
x
x
VxVxVx
V
===
=1220120124040
x
x
x
x
MxMxMx
M
-
=====
=kNRVxkNRVx
VxV
Bx
Bx
x
x
25.5201225.5124
040
===
=020120124040
x
x
x
x
VxVxVx
V
( )( )
====
===
44201244124
40
qVxqVx
xqxVxV
x
x
x
x
====
==
25.1425.5201225.1425.5124
40
x
x
x
x
VxVx
xVxV
( )( )
====
==
2125.5420122125.54124
040
xxRMxxxRMx
MxM
Bx
Bx
x
x
( )( )( )( )
+==+==
==
=842420128424124
2240
22
xxqMxxxqMx
xqxMx
M
x
x
x
x
===
=1220120124040
x
x
x
x
MxMxMx
M
=+==+=
==
=2525.112842125.52012
1325.1842125.512422
4022
xxxMxxxxMx
xqxMx
M
x
x
x
x
-
RA
MA
( )( )( )( ) ( )( ) ( )( ) mkNMRMM
kNRkNkNRF
AAAB
AAVertical
=+==
===
22163440
103240Reactions
@
mkNqmxkNPmx
mkNMkNPxLoading
O
O
/3241
22100
====
===
( )( )( ) 0)4(....6)2(...6)1(....10)1(....100
3122/3642641021
1010
=====
==
=
+ VVVVV
xxmkNkNxkNkNkNx
kNxV
( )( )( )
( ) 0)4(....6)2(....12)1(....2205.1122425.161842
61814102221102210
22
====
+=++=+
+=
MMMMxxxxx
xxxxxx
M
-
The structure shown is constructed of a W10x112 rolled-steel
beam. Draw the shear and bending-moment diagrams for the beam and
the given loading.
Replace the 10 kip load with an equivalent force-couple system
at D. Find the reactions at B by considering the beam as a rigid
body.
Solution
( )( ) kipsRRF BB 34010380 ====( )( )( ) ( )( )
ftkipsMMM
B
B
=
=++==318
01052012380
Vdx
dMqdxdV
==
( )
( )2
40
42
10
1
5.100
5.1
300
33
xMCM
CxM
xVCV
CxVqdxdV
x
x
x
===
+=
===
+===
( )
( ) 9624969624
242424
0
58
5
28
2
+=+==+=
===
==
xMCMCxM
VCV
CVdxdV
x
x
( )
( )
( )
( ) ( ) 226342262016834
343434
240
611
6
311
113
+==+=
+=
===
===
+
+
xMCMCxM
VCV
VCVdxdV
x
x
80 x 118 x 1611 x
-
( )
( ) 34)16(....34)11(...24)11(....24)8(....003410241611
2483118380
=====
==
=
+ VVVVV
xx
xxV
( )( ) ( )
( ) 31816...148)11(....168)11(....96)8(1110204241611
4241185.180 2
====
+
=
+ MMMM
xxxxx
xxM
-
Draw the shear and bending moment diagrams for the beam
shown.
Vdx
dMqdxdV
==
( )
( ) axwxwMCM
CaxwxwM
axwxwVCV
CaxwxwV
axwww
dxdV
ox
ox
oox
oox
oox
6200
62
200
2
320
20
2
320
2
10
1
2
+===
++=
+===
++=+==
ax 0a
xwww oox =
==+
==
=+==
320
320
220
@aLawMMaLawM
awRRawF
oCC
oC
oCC
oVertical
RC
MA
3
2
2awMawV oBoB ==
-
( )
( ) 6263
2
222
0
22
4
2
4
3
3
awxawMawCawM
CxawM
awVawCawV
CVdxdV
oox
ooa
ox
ox
ooa
+=+==
+=
===
==
Lxa
62
2
2awaLwMawV ooCoC +==
-
Shear and Bending Stresses in Beams
Beams under pure bending
constant0 == Mdx
dM
Four Point Bending (Pure bending in the central region)
-
is the radius of curvature
dsd
dds
==
=1
Curvature of a Beam
For very small deflections ds=dx
dxd
== 1
Sign convention for the curvature
-
Deformations of a beam in pure bending: (a) a side view of beam,
(b) cross section beam, and (c) deformed beam. To evaluate normal
strains, consider
the line ef and assumed that the x-axis is the neutral axis of
the undeformed beam.
The length L1 of the line ef after bending takes place:
( ) dxydxdyL
==1
ddxdxd
==1
As the original length of ef is dx, then the elongation is
ydxydxL x ===1
For y positive there is a shortening of the length, that is a
negative eleongation.
-
Normal Stresses in Beams (linearly elastic materials)dxyx
=
EyE xx ==
The stresses acting on the cross section vary linearly with the
distance.
( ) ( )( ) ( )vevey
vevey
x
x
++
It is necessary to locate the origin of the coordinates, that is
the neutral axis of the cross section. Note:(a)The resultant force
in the x-direction is zero (pure bending).(b)The resultant moment
is equal to the bending moment M.
(a) The resultant force in the x-direction is zero (pure
bending).
==== 00 ydAydAEEydAdAx
This means that the first moment of the area of the cross
section (evaluated with respect to the z-axis) is zero. The z-axis
must pass through the centroid of the cross section.
-
(b) The resultant moment is equal to the bending moment M.
====AAA
xx dAyEdAyEdAyMdAydM 22
12 == EI
MIdAyA
The element of moment acts opposite in direction to the positive
bending moment M.Moment-Curvature Equation
EI is called the Flexural Rigidity (measure of the resistance of
a beam to bending)
Relationships between signs of bending moments and signs of
curvatures.
EyEy x
x
== 1
EyEIM x
==
1
IMy
x = Flexure formula. It calculate the bending stresses or
flexural stresses
-
Please Note:Consider the points m1 and m2 located at a distance
x and x + x from the origin respectively. Point m1 has a deflection
equal to and point m2 has a deflection equal to + ,where is the
increment in deflection as we move from m1to m2.
The angle of rotation is for point m1 and + for point m2 , i.e.,
the angle between the lines normal to the tangents at points m1 and
m2 is . The point of intersection of these normals is the center of
curvature O and the distance O to m1 is the radius of curvature
.
-
dsddds
=== 1
dsdxCos
dsdSin
dxd
==
==
tanCurve Deflection theof Slope
If the angle of rotation is very small then s ~ x ; the
curvature = 1/ = / x ; and tan ~ = /x. dx
ddxd
== 1
2
2
xx
= EyEIM x
==
12
21xEI
M
===
EIM
x=2
2
Where M is the bending moment and EI is the flexural rigidity of
the beam. This equation is known as the differential equation of
the deflection
curve.
This equation can be integrated in each particular case to find
the deflection , provided the bending moment M and flexural
rigidity EI are known as functions of x.
-
2
22
1
11
SM
IMc
SM
IMc
==
==
S1 and S2 are known as the section moduli.
Maximum stresses
612
23 bhSbhI ==
3264
34 dSdI ==
Basic Assumptions
Doubly Symmetric Shapes
Beam is slender; plane xy is a plane of symmetry where the load
is applied and the deflections takes place.The thickness remains
constant.The axis of the beam passes through the centroid.
