Fourth Edition
CHAPTER
MECHANICS OF
MATERIALS
Ferdinand P. Beer, E. Russell Johnston, Jr., John T. DeWolf
Lecturer: Nazarena Mazzaro, Ph.D.
Aalborg University
Denmark
Analysis and Design
of Beams for Bending
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Analysis and Design of Beams for Bending
Part 1: 45 min
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
Nazarena Mazzaro, AAU
Part 2: 30 min
Design of Prismatic Beams for
Bending
Sample Problem 5.8
Singularity Functions
Example 5.05
Exercises
5-2
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Introduction
• Objective - Analysis and design of beams
• Beams - structural members supporting loads at
various points along the member
• Transverse loadings of beams are classified as
concentrated loads or distributed loads
• 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
σx = −
My
I
σm =
Mc M
=
I
S
Elastic Flexure
Formulas
Requires determination of the location and
magnitude of largest bending moment
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5-3
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Beer • Johnston • DeWolf
Introduction
Classification of Beam Supports
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5-4
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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 V and V’
and bending couples M and M’ -> positive
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5-5
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 5.01
Draw the shear and bending-moment 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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5-6
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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:
∑ Fy = 0; R − V = 0;V = R = 1 / 2 P
∑ M = 0;− R x + M = 0; M = R x = Px / 2
A
D
A
A
A
∑ Fy = 0; R + V = 0;V = − R = −1 / 2 P
∑ M = R ( L − x) − M = 0; M = P( L − x) / 2
B
E
B
B
V is constant between concentrated loads and M
varies linearly
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.1
SOLUTION:
• Treating the entire beam as a rigid
body, determine the reaction forces
For the timber beam and loading
shown, draw the shear and bendmoment diagrams and determine the
maximum normal stress due to
bending.
• Section the beam at points near
supports and load application points.
Apply equilibrium analyses on
resulting free-bodies to determine
internal shear forces and bending
couples
• Identify the maximum shear and
bending-moment from plots of their
distributions.
• Apply the elastic flexure formulas to
determine the corresponding
maximum normal stress.
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5-8
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.1
SOLUTION:
• Treating the entire beam as a rigid body
∑ Fy = 0 = −20 + R − 40 + R ⇒ R =60 − R
∑ M = 0 = 20 × 2,5 − 40 × 3 + R × 5 ⇒ R = 14
B
B
D
D
D
B
D
RD = 14; RB = 46
• Section the beam and apply equilibrium analyses
on resulting free-bodies
∑ Fy = 0
− 20 kN − V1 = 0
V1 = −20 kN
∑ M1 = 0
(20 kN )(0 m ) + M1 = 0
M1 = 0
∑ Fy = 0
− 20 kN − V2 = 0
V2 = −20 kN
∑ M2 = 0
(20 kN )(2.5 m ) + M 2 = 0
M 2 = −50 kN ⋅ m
V3 = +26 kN
M 3 = −50 kN ⋅ m
V4 = +26 kN M 4 = +28 kN ⋅ m
V5 = −14 kN
M 5 = +28 kN ⋅ m
V6 = −14 kN M 6 = 0
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.1
• Identify the maximum shear and bendingmoment from plots of their distributions.
Vm = 26 kN M m = M B = 50 kN ⋅ m
• Apply the elastic flexure formulas to
determine the corresponding
maximum normal stress.
S = 16 b h 2 = 16 (0.080 m )(0.250 m )2
= 833.33 × 10− 6 m3
MB
50 × 103 N ⋅ m
σm =
=
S
833.33 × 10− 6 m3
V1=-20, M1=0; V2=-20, M2=-50; V3= 26, M3=-50;
V4=26, M4= 28; V5=-14, M5=28, V6=-14, M6=0
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σ m = 60.0 ×106 Pa
5 - 10
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.2
SOLUTION:
• 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.
• Section the beam at points near the
support and load application points.
Apply equilibrium analyses on
The structure shown is constructed of a
resulting free-bodies to determine
W10x112 rolled-steel beam. (a) Draw
internal shear forces and bending
the shear and bending-moment diagrams
couples.
for the beam and the given loading. (b)
determine normal stress in sections just
• Apply the elastic flexure formulas to
to the right and left of point D.
determine the maximum normal
stress to the left and right of point D.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.2
SOLUTION:
• Replace the 10 kip load with equivalent forcecouple system at D. Find reactions at B.
• Section the beam and apply equilibrium
analyses on resulting free-bodies.
From A to C :
∑ Fy = 0
∑ M1 = 0
− 3x − V = 0
(3x )(12 x )+ M
From C to D :
∑ Fy = 0 − 24 − V = 0
V = −3 x kips
= 0 M = −1.5 x 2 kip ⋅ ft
V = −24 kips
∑ M 2 = 0 24( x − 4) + M = 0 M = (96 − 24 x ) kip ⋅ ft
From D to B :
V = −34 kips
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M = (226 − 34 x ) kip ⋅ ft
5 - 12
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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 X-X axis.
To the left of D :
M 2016 kip ⋅ in
=
S
126 in 3
To the right of D :
σ m = 16.0 ksi
M 1776 kip ⋅ in
=
S
126 in 3
σ m = 14.1 ksi
σm =
σm =
• Concentrated loads: V constant
M varies linearly
• Distributed load: V varies linearly
M parabola
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Relations Among Load, Shear, and Bending Moment
• Relationship between load and shear:
∑ Fy = 0 : V − (V + ∆V ) − w ∆x = 0
∆V = − w ∆x
dV
= −w
dx
xD
VD − VC = − ∫ w dx
xC
• Relationship between shear and bending
moment:
∑ M C′ = 0 :
(M + ∆M ) − M − V ∆x + w∆x ∆x = 0
2
1
∆
M
2
∆M = V ∆x − 12 w (∆x ) ⇒
= V − w∆x
2
∆x
∆x → 0;
dM
=V
dx
M D − MC =
xD
∫ V dx
xC
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5 - 14
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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:
x
V − VA = − ∫ w.dx = − wx
0
V = VA − wx =
1
1
wL − wL = w( L − x)
2
2
x
M − M A = ∫ V .dx; M A = 0
0
wL2
1
1
2
M = ∫ w( L − x)dx = w( Lx − x ) ⇒ M max
2
2
8
0
x
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.3
SOLUTION:
• Taking the entire beam as a free body,
determine the reactions at A and D.
• Apply the relationship between shear and
load to develop the shear diagram.
Draw the shear and bending
moment diagrams for the beam
and loading shown.
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• Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
5 - 16
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.3
SOLUTION:
• Taking the entire beam as a free body, determine the
reactions at A and D.
∑MA = 0
0 = D(24 ft ) − (20 kips )(6 ft ) − (12 kips )(14 ft ) − (12 kips )(28 ft )
D = 26 kips
∑ Fy = 0
0 = Ay − 20 kips − 12 kips + 26 kips − 12 kips
Ay = 18 kips
• Apply the relationship between shear and load to
develop the shear diagram.
dV
= −w
dx
dV = − w dx
- zero slope between concentrated loads
- linear variation over uniform load segment
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.3
• Apply the relationship between bending
moment and shear to develop the bending
moment diagram.
dM
=V
dx
dM = V dx
- bending moment at A and E is zero
- bending moment variation between A, B,
C and D is linear
- bending moment variation between D
and E is quadratic
- net change in bending moment is equal to
areas under shear distribution segments
D
M D − M C = ∫ V .dx
C
- total of all bending moment changes across
the beam should be zero (108-16-140+48=0)
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.5
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.
Draw the shear and bending moment
diagrams for the beam and loading
shown.
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• Apply the relationship between
bending moment and shear to develop
the bending moment diagram.
5 - 19
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.5
SOLUTION:
• Taking the entire beam as a free body,
determine the reactions at C.
∑ Fy = 0 = − 12 w0 a + RC
a⎞
⎛
∑ M C = 0 = 12 w0 a⎜ L − ⎟ + M C
3⎠
⎝
RC = 12 w0 a
a⎞
⎛
M C = − 12 w0 a⎜ L − ⎟
3⎠
⎝
Results from integration of the load and shear
distributions should be equivalent.
• Apply the relationship between shear and load
to develop the shear diagram.
a
2 ⎞⎤
⎡ ⎛
x
⎛ x⎞
VB − V A = − ∫ w0 ⎜1 − ⎟ dx = − ⎢ w0 ⎜ x − ⎟⎥
⎜
a⎠
2a ⎟⎠⎥⎦
⎢⎣ ⎝
0 ⎝
0
a
VB = − 12 w0 a = − ( area under load curve)
- No change in shear between B and C.
- Compatible with free body analysis
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.5
• Apply the relationship between bending moment
and shear to develop the bending moment
diagram.
a
⎡
⎛ x 2 x3 ⎞⎤
⎛
x 2 ⎞⎟ ⎞⎟
⎜
⎜
M B − M A = ∫ − w0 x −
dx = ⎢− w0 ⎜ − ⎟⎥
⎜ 2 6a ⎟
⎜
⎜
2a ⎟⎠ ⎟⎠
⎢⎣
0⎝
⎠⎥⎦ 0
⎝
⎝
a⎛
M B = − 13 w0 a 2
L
(
)
M B − M C = ∫ − 12 w0 a dx = − 12 w0 a(L − a )
a
a w0 ⎛
a⎞
M C = − 16 w0 a(3L − a ) =
⎜L− ⎟
2 ⎝
3⎠
Results at C are compatible with free-body
analysis
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Design of Prismatic Beams for Bending
• The largest normal stress is found at the surface where the
maximum bending moment occurs.
M max c M max
σm =
=
I
S
• 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.
σ m ≤ σ all
S min =
M max
σ all
• 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.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Design of Prismatic Beams for Bending
• Determine σall (If σall is the same for tension and compresion then
follow 1,2,3 – Otherwise consider 4 in step 2)
• 1-Draw shear and bending-moment diagrams and determine |M|max
• 2-Determine Smin
• 3-Find the dimentions of the beam to use: b, h for S>Smin
• 4-Select the beam section so that σm<= σall for tensile and
compressive stresses.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.7
60 kN/m
2.4 m
20 kN/m
1.2 m
90 mm
SOLUTION:
• Considering the entire beam as a freebody, determine the reactions at A and
B.
• Develop the shear diagram for the
beam and load distribution. From the
diagram, determine the maximum
bending moment.
A 3.6 m-long overhanging timber
beam AC is to be designed to support
the distributed and concentrated loads
shown. Knowing that timber of 100mm nominal width (90-mm actual
• Determine the minimum acceptable
width) with a 12 MPa allowable stress
beam section modulus. Find the value
is to be used, determine the minimum
for the h.
required depth h of the beam.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 5.7
14.4 kN
• Considering the entire beam as a free-body
20 kN
∑M
A
= 0 = B(2.4 m ) − (14.4 kN )(1.2 m ) − (20 kN )(3.6 m )
B = 37.2 kN = Vc
∑F
1.2m
2.4 m
y
= 0 = Ay + 37.2kN − 14.4 kN − 20 kN
Ay = −2.8 kN = VA
20
kN
(+24)
• Plot shear diagram and determine Mmax.
VB − VA = −(area under load curve) = −(6kN / m)(2.4m) = −14.4kN
VB = VA − 14 − 4 = −2.8kN − 14.4kN = −17.2kN
VB = −17.2 kN
-2.8
kN
(+24)
-17.2 kN
• MA=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
|M|max= 24 kN.m
S min =
M
max
σ all
=
24kN .m
= 2 ×10 6 mm 3
12 MPa
1 2
1
bh ≥ S min ⇒ (90mm)h 2 ≥ 2 ×10 6 mm 3 ⇒ h ≥ 365.2mm
6
6
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