MECHANICS OF
MATERIALS
Fourth Edition
Ferdinand P. Beer, E. Russell Johnston, Jr., John T. DeWolf
CHAPTER
Analysis and Design
of Beams for Bending
Lecturer: Nazarena Mazzaro, Ph.D.
Aalborg University
Denmark
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Analysis and Design of Beams for Bending
Introduction
Shear and Bending Moment
Diagrams
Example 5.01
Sample Problem 5.1
Sample Problem 5.2
Relations Among Load, Shear, and
Bending Moment
Example 5.03
Sample Problem 5.3
Sample Problem 5.5
Part 1: 45 min
Design of Prismatic Beams for
Bending
Sample Problem 5.8
Singularity Functions
Example 5.05
Part 2: 30 min
Exercises
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Introduction
•Beamsstructural members supporting loads at
various points along the member
•Objective Analysis and design of beams
•Transverse loadings of beams are classified as
concentratedloads or distributedloads
•Applied loads result in internal forces consisting
of a shear force(from the shear stress
distribution) and a bending couple(from the
normal stress distribution)
•Normal stress
is often the critical design criteria
S
M
I
cM
I
My
mx
==−=
σσ
Requires determination of the location and
magnitude of largest bending moment
Elastic Flexure
Formulas
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Introduction
Classification of Beam Supports
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Shear and Bending Moment Diagrams
•Determination of maximum normal and
shearing stresses requires identification of
maximum internal shear force V and
bending couple M.
•Shear force and bending couple at a point are
determined by passing a section through the
beam and applying an equilibrium analysis on
the beam portions on either side of the
section.
•Sign conventions for shear forces Vand V’
and bending couples Mand M’> positive
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Example 5.01
Draw the shear and bendingmoment diagrams for a simply
supported beam AB of span L subjected to a single concentrated
load P at it midpoint C
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Example 5.01
•Determination of the reactions RA=RB=1/2P
•Cut the beam at D and plot free body diagrams with
positive V and M. Equilibrium equations:
2/)(;0)(
2/1;0;0
2/;0;0
2/1;0;0
xLPMMxLRM
PRVVRFy
PxxRMMxRM
PRVVRFy
B
E
BB
AAD
AA
−==−−=
−=−==+=
===+−=
===−=
∑
∑
∑
∑
V is constant between concentrated loads and M
varies linearly
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Sample Problem 5.1
For the timber beam and loading
shown, draw the shear and bend
moment diagrams and determine the
maximum normal stress due to
bending.
SOLUTION:
•Treating the entire beam as a rigid
body, determine the reaction forces
•Identify the maximum shear and
bendingmoment from plots of their
distributions.
•Apply the elastic flexure formulas to
determine the corresponding
maximum normal stress.
•Section the beam at points near
supports and load application points.
Apply equilibrium analyses on
resulting freebodies to determine
internal shear forces and bending
couples
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Sample Problem 5.1
SOLUTION:
•Treating the entire beam as a rigid body
•Section the beam and apply equilibrium analyses
on resulting freebodies
()()
00m0kN200
kN200kN200
111
11
==+
∑
=
−
=
=
−
−
∑
=
MMM
VV
F
y
()()
mkN500m5.2kN200
kN200kN200
222
22
⋅−==+
∑
=
−
=
=
−
−
∑
=
MMM
VV
F
y
0kN14
mkN28kN14
mkN28kN26
mkN50kN26
66
55
44
33
=−=
⋅+=−=
⋅+=+=
⋅
−
=
+
=
MV
MV
MV
M
V
46;14
1453405,2200
6040200
==
=⇒×+×−×==
−=⇒+−+−==
∑
∑
BD
DDB
BDDB
RR
RRM
RRRRFy
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Sample Problem 5.1
•Identify the maximum shear and bending
moment from plots of their distributions.
mkN50kN26
⋅
=
=
=
Bmm
MMV
•Apply the elastic flexure formulas to
determine the corresponding
maximum normal stress.
(
)
(
)
36
3
36
2
6
1
2
6
1
m1033.833
mN1050
m1033.833
m250.0m080.0
−
−
×
⋅×
==
×=
==
S
M
hbS
B
m
σ
Pa100.60
6
×=
m
σ
V1=20, M1=0; V2=20, M2
=50; V3= 26, M3=50;
V4=26, M4= 28; V5=14, M5=28, V
6=14, M
6=0
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Sample Problem 5.2
The structure shown is constructed of a
W10x112 rolledsteel beam. (a) Draw
the shear and bendingmoment diagrams
for the beam and the given loading. (b)
determine normal stress in sections just
to the right and left of point D.
SOLUTION:
•Replace the 10 kip load with an
equivalent forcecouple system at D.
Find the reactions at B by considering
the beam as a rigid body.
•Section the beam at points near the
support and load application points.
Apply equilibrium analyses on
resulting freebodies to determine
internal shear forces and bending
couples.
•Apply the elastic flexure formulas to
determine the maximum normal
stress to the left and right of point D.
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Sample Problem 5.2
SOLUTION:
•Replace the 10 kip load with equivalent force
couple system at D. Find reactions at B.
•Section the beam and apply equilibrium
analyses on resulting freebodies.
()
()
ftkip5.1030
kips3030
:
2
2
1
1
⋅−==+
∑
=
−==−−
∑
=
xMMxxM
xVVxF
C
to
A
From
y
()()
ftkip249604240
kips240240
:
2
⋅−==+−
∑
=
−==−−
∑
=
xMMxM
VVF
D
to
C
From
y
()
ftkip34226kips34
:
⋅−=−=xMV
Bto
D
From
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Sample Problem 5.2
•Apply the elastic flexure formulas to
determine the maximum normal stress to
the left and right of point D.
From Appendix C for a W10x112 rolled
steel shape, S= 126 in3
about the XXaxis.
3
3
in126
inkip1776
:
in126
inkip2016
:
⋅
==
⋅
==
S
M
DofrighttheTo
S
M
D
o
f
le
ft
theTo
m
m
σ
σ
ksi0.16
=
m
σ
ksi1.14
=
m
σ
•Concentrated loads: V constant
M varies linearly
•Distributed load: V varies linearly
M parabola
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Relations Among Load, Shear, and Bending Moment
(
)
xwV
xwVVVFy
∆−=∆
=
∆
−
∆
+
−
=
∑
0:0
∫
−=−
−=
D
C
x
x
CD
dxwVV
w
dx
dV
•Relationship between load and shear:
()
()
xwV
x
M
xwxVM
x
xwxVMMMM
C
∆−=
∆
∆
⇒∆−∆=∆
=
∆
∆+∆−−∆+=
∑
′
2
1
0
2
:0
2
2
1
∫
=−
=→∆
D
C
x
x
CD
dxVMM
V
dx
dM
x;0
•Relationship between shear and bending
moment:
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Example 5.03
Draw the shear and bending moment diagrams for the
simply supported beam and determine the maximum value
of the bending moment.
SOLUTION:
•RA=RB=wL/2
•Determination of V and M at any distance from A:
8
)(
2
1
)
2
1
(
0;.
)
2
1
(
2
1
.
2
max
2
0
0
0
wL
MxLxwdxxLwM
MdxVMM
xLwwLwLwxVV
wxdxwVV
x
A
x
A
A
x
A
⇒−=−=
==−
−=−=−=
−=−=−
∫
∫
∫
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Sample Problem 5.3
Draw the shear and bending
moment diagrams for the beam
and loading shown.
SOLUTION:
•Taking the entire beam as a free body,
determine the reactions at Aand D.
•Apply the relationship between shear and
load to develop the shear diagram.
•Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
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Sample Problem 5.3
SOLUTION:
•Taking the entire beam as a free body, determine the
reactions at Aand D.
()()()()()()()
kips18
kips12kips26kips12kips200
0F
kips26
ft28kips12ft14kips12ft6kips20ft240
0
y
=
−+−−=
=
∑
=
−−−=
=
∑
y
y
A
A
A
D
D
M
•Apply the relationship between shear and load to
develop the shear diagram.
dxwdVw
dx
dV
−=−=
zero slope between concentrated loads
linear variation over uniform load segment
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Sample Problem 5.3
•Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
dxVdMV
dx
dM
==
bending moment at Aand Eis zero
total of all bending moment changes across
the beam should be zero (10816140+48=0)
net change in bending moment is equal to
areas under shear distribution segments
bending moment variation between D
and Eis quadratic
bending moment variation between A, B,
C and Dis linear
dxVMM
D
C
CD
.
∫
=−
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Sample Problem 5.5
Draw the shear and bending moment
diagrams for the beam and loading
shown.
SOLUTION:
•Taking the entire beam as a free body,
determine the reactions at C.
•Apply the relationship between shear
and load to develop the shear diagram.
•Apply the relationship between
bending moment and shear to develop
the bending moment diagram.
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Sample Problem 5.5
SOLUTION:
•Taking the entire beam as a free body,
determine the reactions at C.
⎟
⎠
⎞
⎜
⎝
⎛
−−=+
⎟
⎠
⎞
⎜
⎝
⎛
−==
∑
=+−==
∑
33
0
0
0
2
1
0
2
1
0
2
1
0
2
1
a
LawMM
a
LawM
awRRawF
CC
C
CCy
Results from integration of the load and shear
distributions should be equivalent.
•Apply the relationship between shear and load
to develop the shear diagram.
()
curveloadunderareaawV
a
x
xwdx
a
x
wVV B
a
a
AB
−=−=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
∫
⎟
⎠
⎞
⎜
⎝
⎛
−−=−
0
2
1
0
2
0
0
0
2
1
No change in shear between Band C.
Compatible with free body analysis
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Sample Problem 5.5
•Apply the relationship between bending moment
and shear to develop the bending moment
diagram.
2
0
3
1
0
32
0
0
2
0
622
awM
a
xx
wdx
a
x
xwMM
B
a
a
AB
−=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=
∫
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−=−
(
)
()
()
⎟
⎠
⎞
⎜
⎝
⎛
−=−−=
−−=
∫
−=−
32
3
0
0
6
1
0
2
1
0
2
1
a
L
wa
aLawM
aLawdxawMM
C
L
a
CB
Results at Care compatible with freebody
analysis
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Design of Prismatic Beams for Bending
•Among beam section choices which have an acceptable
section modulus, the one with the smallest weight per unit
length or cross sectional area
will be the least expensive
and the best choice.
•The largest normal stress is found at the surface where the
maximum bending moment occurs.
S
M
I
cM
m
maxmax
==
σ
•A safe design requires that the maximum normal stress be
less than the allowable stress for the material used. This
criteria leads to the determination of the minimum
acceptable section modulus.
al
l
allm
M
S
σ
σ
σ
max
min
=
≤
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Design of Prismatic Beams for Bending
•Determine σall
(Ifσall
is thesame for tension and compresionthen
follow1,2,3 –Otherwiseconsider4 in step 2)
•1Draw shearand bendingmomentdiagrams and determineM
max
•2Determine S
min
•3Find thedimentionsofthebeamto use: b, h for S>Smin
•4Select thebeamsectionso thatσm<= σall for tensileand
compressivestresses.
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Sample Problem 5.7
A 3.6 mlong overhanging timber
beam AC is to be designed to support
the distributed and concentrated loads
shown. Knowing that timber of 100
mm nominal width (90mm actual
width) with a 12 MPaallowable stress
is to be used, determine the minimum
required depth h of the beam.
SOLUTION:
•Considering the entire beam as a free
body, determine the reactions at Aand
B.
•Develop the shear diagram for the
beam and load distribution. From the
diagram, determine the maximum
bending moment.
•Determine the minimum acceptable
beam section modulus. Find the value
for the h.
60 kN/m
20 kN/m
90 mm
2.4 m
1.2 m
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Sample Problem 5.7
•Considering the entire beam as a freebody
(
)
(
)
(
)
(
)
(
)
A
VkN8.2
kN20kN4.14kN2.370
VckN2.37
m6.3kN20m2.1kN4.14m4.20
=−=
−−+==
==
−−==
∑
∑
y
yy
A
A
AF
B
BM
•Plot shear diagram and determine M
max.
(
)
kN2.17
2.174.148.2414
4.14)4.2)(/6(
−=
−=−−=−−=
−
=
−
=
−
=
−
B
AB
AB
V
kNkNkNVV
kNmmkNcurveloadunderareaVV
•M
A=Mc=0, Between A and B, M decreases an
amount equal to area VAB, and between B and
C in increases the same amount. Thus the
Mmax= 24 kN.m
mmhmmhmmSbh
mm
MPa
mkN
M
S
all
2.365102)90(
6
1
6
1
102
12
.24
362
min
2
36
max
min
≥⇒×≥⇒≥
×===
σ
14.4 kN
20 kN
2.4 m
1.2m
20
kN
(+24)
(+24)
17.2 kN
2.8
kN
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