SHEAR FOR
SHEAR FOR
CE &
CE &
BENDING
BENDING
MOMENT DIAGRAM
V
V
arious T
arious T
ypes
ypes
of
of
Beam
Beam
Loading
Loading
&
&
Suppor
Suppor
t
t
•
• Beam Beam - - structural membstructural member designed to er designed to supportsupport
loads applied at various points along its length. loads applied at various points along its length.
•
• Beam design Beam design is two-step process: is two-step process:
1)
1) deterdetermine smine shearihearing forng forces and bces and bendingending moments produced by applied loads moments produced by applied loads 2)
2) selecselect cross-st cross-sectioection best suin best suited to resited to resistst shearing forces and bending moments shearing forces and bending moments
•
• Beam can be subjected toBeam can be subjected to concentrated concentrated loads or loads or
distributed
distributed loads or combination of both. loads or combination of both.
Beer,
Various Types of Beam Loading and Support
• Beams are classified according to way in which they are
supported.
• Reactions at beam supports are determinate if they
involve only three unknowns. Otherwise, they are statically indeterminate.
Types of Beams
•
Depends on the support configuration
M
F
vF
HFixed
F
VF
VF
HPin
Roller
Pin
Roller
F
VF
VF
HStatically Indeterminate Beams
Continuous Beam
Propped Cantilever Beam
SHEAR FORCES AND BENDING MOMENTS
•
When a beam is loaded by forces or
couples, stresses and strains are
created throughout the interior of the
beam.
•
To determine these stresses and
strains, the internal forces and
internal couples that act on the cross
sections of the beam must be found.
Shear and Bending Moment in a Beam
• Wish to determine bending moment
and shearing force at any point in a beam subjected to concentrated and
distributed loads.
• Determine reactions at supports by
treating whole beam as free-body.
• Cut beam at C and draw free-body
diagrams for AC and CB. By
definition, positive sense for internal force-couple systems are as shown.
• From equilibrium considerations,
determine M and V or M’ and V’ .
Shear and Bending Moment Diagrams
• Variation of shear and bending
moment along beam may be plotted.
• Determine reactions at
supports.
• Cut beam at C and consider
member AC ,
2 2 M Px P
V
• Cut beam at E and consider
member EB,
2 2 M P L x PV
• For a beam subjected to
concentrated loads, shear is
constant between loading points and moment varies linearly.
Sample Problem
Draw the shear and bending moment diagrams for the beam and loading shown.
SOLUTION:
• Taking entire beam as a free-body,
calculate reactions at B and D.
• Find equivalent internal force-couple
systems for free-bodies formed by cutting beam on either side of load application points.
• Plot results.
Sample Problem
SOLUTION:
• Taking entire beam as a free-body, calculate
reactions at B and D.
• Find equivalent internal force-couple systems at
sections on either side of load application points.
F y
0: 20kN 0 1
V V 1
20kN : 0 2
M
20kN
0m
M 1
0 M 1
0 m kN 50 kN 26 m kN 50 kN 26 m kN 50 kN 26 m kN 50 kN 26 6 6 5 5 4 4 3 3
M V M V M V M V Similarly,Sample Problem
• Plot results.
Note that shear is of constant value between concentrated loads and bending moment varies linearly.
Sample Problem 2
Draw the shear and bending moment diagrams for the beam AB. The
distributed load of 40 lb/in. extends over 12 in. of the beam, from A to C , and the 400 lb load is applied at E .
SOLUTION:
• Taking entire beam as free-body,
calculate reactions at A and B.
• Determine equivalent internal
force-couple systems at sections cut within segments AC , CD, and DB.
• Plot results.
Sample Problem
SOLUTION:
• Taking entire beam as a free-body, calculate
reactions at A and B. :
0
M A
32in.
480lb
6in.
400lb
22in.
0y B lb 365
y B : 0
M B
480lb
26in.
400lb
10in.
A 32in.
0 lb 515
A : 0
F x B x
0• Note: The 400 lb load at E may be replaced by a
400 lb force and 1600 lb-in. couple at D.
Sample Problem
: 0 1
M
515 x
40 x 21 x
M
0 2 20 515 x x M
: 0 2
M
515 x
480
x
6
M
0
2880
35
lb
in.
x M From C to D:
F y
0: 515
480
V
0 lb 35
V• Evaluate equivalent internal force-couple systems
at sections cut within segments AC , CD, and DB.
From A to C :
F y
0: 515
40 x
V
0x V
515
40Sample Problem
: 0 2
M
6
1600 400
18
0 480 515
x x x M
11,680
365
lb
in.
x M• Evaluate equivalent internal force-couple
systems at sections cut within segments AC , CD, and DB. From D to B:
F y
0: 515
480
400
V
0 lb 365
VSample Problem
• Plot results. From A to C : x V
515
40 2 20 515 x x M
From C to D: lb 35
V
2880
35
lb
in.
x M From D to B: lb 365
V
11,680
365
lb
in.
x MRelations Among Load, Shear, and Bending Moment
• Relations between load and shear:
w x V dx dV x w V V V x
lim0 0
area underloadcurve
D C x x C D V wdx V• Relations between shear and bending moment:
V w x
V x M dx dM x x w x V M M M x x
2 1 0 0 lim lim 0 2
area undershearcurve
D C x x C D M V dx M• Reactions at supports, 2 wL R R A
B
• Shear curve,
x L w wx wL wx V V wx dx w V V A x A 2 2 0 • Moment curve,
0 at 8 2 2 2 max 2 0 0 V dx dM M wL M x x L w dx x L w M Vdx M M x x ABeer, Ferdinand & Johnston, Russel E. “Vectors
Sample Problem 3
Draw the shear and bending-moment diagrams for the beam and loading shown.
SOLUTION:
• Taking entire beam as a free-body, determine
reactions at supports.
• With uniform loading between D and E , the
shear variation is linear.
• Between concentrated load application
points, and shear is constant. 0
w dx dV• Between concentrated load application
points, The change in moment between load application points is equal to area under shear curve between
points. . constant
V dx dM• With a linear shear variation between D
and E , the bending moment diagram is a
Sample Problem 3
• Between concentrated load application points,
and shear is constant.
0
w dx dV• With uniform loading between D and E , the shear
variation is linear. SOLUTION:
• Taking entire beam as a free-body,
determine reactions at supports.
M A
0:
12kips
28ft 0 ft 14 kips 12 ft 6 kips 20 ft 24
D kips 26
D : 0
F y
0 kips 12 kips 26 kips 12 kips 20
y A kips 18
y ASample Problem
• Between concentrated load application
points, The change in moment between load application points is equal to area under the shear curve between points. . constant
V dx dM• With a linear shear variation between D
and E , the bending moment diagram is a parabola. 0 48 ft kip 48 140 ft kip 92 16 ft kip 108 108
