Beams
Beams:
Comparison with
trusses, plates
Examples:
1. simply supported beams
2. cantilever beams
L, W, t: L >> W and L >> t
L
W
t
Beams

loads and internal loads
Loads: concentrated loads, distributed loads, couples (moments)
Internal loads:
shear force
and
bending moments
q
q
> 0
q
< 0
Shear Forces, Bending Moments

Sign
Conventions
Shear forces
:
Bending moments
:
left section
right section
positive shear:
negative shear:
positive moment
negative moment
Shear Forces, Bending Moments

Static
Equilibrium Approach
Procedure:
1. find reactions;
2. cut the beam at a certain cross section, draw F.B.D. of one piece of the beam;
3. set up equations;
4. solve for shear force and bending moment at that cross section;
5. draw shear and bending moment diagrams.
Example 1: Find the shear force and bending diagram at any cross section of
the beam shown below.
Relationship between Loads, Shear
Forces, and Bending Diagram
dV dM
q V
dx dx
Beam

Normal Strain
Pure bending problem
no transverse load
no axial load
no torque
Observations of the deformed beam under pure bending
Length of the longitudinal elements
Vertical plane remains plane after deformation
Beam deforms like an arc
M
M
Normal Strain

Analysis
x
neutral axis (N.A.):
radius of curvature:
Coordinate system:
longitudinal strain:
y
q
r
'
x
L x x
L x
y
y
r q rq
rq r
N.A.
Beam

Normal Stress
x x
Ey
E
r
Hooke’s Law:
Maximum stresses:
M
M
M
x
y
Neutral axis:
0 0 0
0 0
x x
A A
c
A
Ey
F dA dA
ydA y
r
Flexure Formula
M M
2
2
1
x
A A
y E M
dM dA y M E dA y dA
EI
r r r
2
: second moment of inertial (with respe
ct to the neutral axis)
A
I y dA
M
x
y
Moment balance:
x x
Ey
E
r
x
M y
I
Axially loaded members Torsional shafts:
Comparison:
Moment of Inertia

I
dA
y
I
A
2
Example 2:
Example 3:
w
h
4h
w
w
w
h
Design of Beams for Bending Stresses
Design Criteria:
n
u
Y
allowable
allowable
or
,
1.
2. cost as low as possible
Design Question:
Given the loading and material, how to choose the shape and the size
of the beam so that the two design criteria are satisfied?
Design of Beams for Bending Stresses
Procedure:
•
Find M
max
•
Calculate the required section modulus
•
Pick a beam with the least cross

sectional area or weight
•
Check your answer
; : section modulus
x
M y M I
S
I S y
Design of Beams for Bending Stresses
Example 4: A beam needs to support a uniform loading with density of
200 lb /ft. The allowable stress is 16,000 psi. Select the shape and the size
of the beam if the height of the beam has to be 2 in and only rectangular and
circular shapes are allowed.
6 ft
Shear Stresses inside Beams
shear force: V
Horizontal shear stresses:
V
2
1
t
H
1
1
, : first moment
h
H
y
VQ
Q ydA
Iw
t
x
y
h
1
h
2
y
1
Shear Stresses inside Beams
H
VQ
Iw
t t
Relationship between the horizontal shear stresses and the vertical shear stresses:
Shear stresses

force balance
Iw
VQ
t
V
: shear force at the transverse cross section
Q
: first moment of the cross sectional area above the level at which
the shear stress is being evaluated
w
: width of the beam at the point at which the shear stress is being
evaluated
I
: second moment of inertial of the cross section
x
h
1
h
2
y
1
y
1
1
h
y
Q ydA
Shear Stresses inside Beams
2
L
4
L
Example 5: Find shear stresses at points A, O and B located at cross section
a

a.
P
a
a
4
h
4
L
4
h
4
h
4
h
w
A
B
O
Shear Stress Formula

Limitations
Iw
VQ
t

elementary shear stress theory
Assumptions:
1. Linearly elastic material, small deformation
2. The edge of the cross section must be parallel to y axis, not applicable for
triangular or semi

circular shape
3. Shear stress must be uniform across the width
4. For rectangular shape,
w
should not be too large
Shear Stresses inside Beams
Example 6: The transverse shear V is 6000 N. Determine the vertical shear stress
at the web.
Beams

Examples
Example 7: For the beam and loading shown, determine
(1) the largest normal stress
(2) the largest shearing stress
(3) the shearing stress at point a
Deflections of Beam
1
M
EI
r
2
2
3 2
2
1
1
d y
dx
dy
dx
r
Deflection curve of the beam: deflection of the neutral axis of the beam.
x
y
x
y
Derivation:
2 2
2 2
d y M d y
EI M
dx EI dx
2
2
dM d d y
V V EI
dx dx dx
Moment

curvature relationship:
Curvature of the deflection curve:
Small deflection:
(1)
(2)
(3)
Equations (1), (2) and (3) are totally equivalent.
P
Deflections by Integration of the
Moment Differential Equation
Example 8 (approach 1):
Deflections by Integration of the Load
Differential Equation
Example 8 (approach 2):
Method of Superposition
P
q
P
Deflection:
y
Deflection:
y
1
Deflection:
y
2
1 2
y y y
Method of Superposition
Example 9
P
F
F
q
A
B
B
C
+
d
B
Method of Superposition
Example 9
P
F
A
B
+
d
B
𝜃
=
𝜃
bending
+
𝜃
𝛿
𝐵
𝜃
bending
𝜃
𝛿
𝐵
𝜃
bending
=
4𝑃
2
81 𝐼
𝜃
𝛿
𝐵
≈
tan
𝜃
𝛿
𝐵
=
𝛿
Method of Superposition
Example 9
F
q
B
C
C
B
C
F
q
𝛿
=
𝛿
𝐹
+
𝛿
𝑞
𝛿
𝐹
=
3
3 𝐼
𝛿
𝑞
=
𝑞
4
8 𝐼
𝜃
A
=
4𝑃
2
81 𝐼
+
2
𝑃
3
9
𝐼
+
𝑞
4
8
𝐼
Statically Indeterminate Beam
Number of unknown reactions is larger than the number of independent
Equilibrium equations.
Propped cantilever beam
Clamped

clamped beam
Continuous beam
Statically Indeterminate Beam
Example 10. Find the reactions of the propped beam shown below.
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