OUTLINE
CHAPTER 7:
Flexural Members
-Types of beams, loads and reactions
-Shear forces and bending moments
-Shear force and bending moment diagrams
-Bending deformation of a straight member
-The flexure formula
-The elastic curve
-Slope and deflection by direct integration method
1
Types of beams
2
Types of loads
Beams are usually described by the manner in which they are supported.
Distributed load
Concentrated load Moment
Simply supported beam Cantilevered beam Overhanging beam
Pin support prevents translation but does not prevent rotation.
Roller support prevents translation only in the vertical direction.
Fixed support prevents translation and rotation.
[Reference # 3: Page 255 – 281]
3
Reactions
The first step in the analysis of beam is finding the reactions.
From the reactions, the shear force and bending moments can be found.
For the statically determinate beams, all reactions can be found from
free-body diagrams and equilibrium equations.
Distributed load
Reaction
4
Shear forces and bending moments
P The stress resultants in statically
determinate beams can be calculated
from equilibrium equations.
P
M V
Concentrated load Moment
V
x
M
∑F
y
= 0, P − V = 0
P =V
∑ M = 0, M − Px = 0
M = Px
Reaction
5
6
Sign convention
V
V
A positive shear force acts
clockwise against the material.
M M
A positive bending moment
compresses the upper part of
the beam.
M V
P
M+M1
dx V+V
1
M0
M+M1
M V
dx V+V
1
7
Homework # 21: Reference # 3 Problem 6-3
Shear force and bending
moment diagrams
P
Relationship between loads, shear
forces and bending moments
q
From equilibrium equations,
M V
M+dM
dV
dM
= −q
=V
dx V+dV
dx
dx
⎛ dx ⎞
V1 = − P M 1 = P⎜ ⎟ + Vdx + V1dx
⎝ 2 ⎠
M 1 = −M 0
8
Shear force and bending
moment diagrams
P
q
P P2 3
q
1
SFD
SFD
BMD
BMD
9
P1
Shear force and bending
moment diagrams
P2
10
Draw the SFD and BMD
q
P
P
b
L
SFD
M1
L/3
BMD
11
b
M1=PL/4
P
L/2
L/2
q0
2M1
L/3
L/3
L
Homework # 22: Reference # 3 Problem 6-30
12
Bending deformation of a
straight member
Bending deformation of a
straight member
Compression
Tension
Need
Reinforcement!
x
z
SFD
BMD
Δs = Δx
Undeformed element
Δs '− Δs
Δs →0
Δs
(
ρ − y )Δθ − ρΔθ
∈= lim
Δθ →0
ρΔθ
y
∈= −
ρ
ρ
Δθ
Δx
∈= lim
longitudinal
axis
Δs’
y
Bending deformation of a
straight member
“Longitudinal normal strain will vary linearly with y from NA”
− ∈max
∈
− y/
⎛ y⎞
c
∈= −⎜ ⎟ ∈max
=
⎝c⎠
y
∈max
c/
longitudinal
axis
ρ
Δx
14
o’
y
longitudinal
axis
neutral surface
13
Bending deformation of a
straight member
Δx
Δx
Normal strain distribution
15
16
The flexure formula
x
σ max
M
σ
The flexure formula
∈max
c
y
FR = ∑ Fx ; 0 = ∫AdF = ∫AσdA
⎛ y⎞
= ∫ − ⎜ ⎟σ max dA
A
⎝c⎠
=
c
∫
A
Bending stress variation
σ max y
⎛ y⎞
⎝c⎠
σ = −⎜ ⎟σ max
Normal strain variation
y
x
⎛ y⎞
∈= −⎜ ⎟ ∈max
⎝c⎠
Deformed element
ρ
y
y
z
Reinforcement
[Reference # 3: Page 282 – 303]
∈
x
Δx
neutral axis
longitudinal
axis
M
x
M+
y
y
z
− σ max
c
∫
A
ydA
M
σ
y
x
Bending stress variation
(M R )z = ∑ M z ; M = ∫A ydF = ∫A y(σdA)
⎞
⎛y
= ∫ y⎜ σ max ⎟ dA
A
⎠
⎝c
ydA = 0
17
σ max =
c
M=
σ max
c
∫
A
y 2 dA
σ =−
Mc
I
My
I
I = The moment of
inertia of the crosssectional area
about NA
18
Example
Example
If the beam has a square cross section of 100 mm on each side,
determine the absolute maximum bending stress in the beam.
1.2 kN
If the beam has a rectangular cross section with a width of 200
mm and a height of 400 mm, determine the absolute maximum
bending stress in the beam.
8 kN/m
A
8 kN
B
0.8 m
10 kN
2 kN/m
40 kN·m
400 mm
0.8 m
4m
6m
200 mm
19
The elastic curve
Sign convention
The elastic curve = the deflection diagram of the longitudinal
axis that passes though the centroid of each cross-sectional
area of the beam.
Force
Displacement
Moment
Rotation or slope
+M
+M
A positive bending moment
compresses the upper part of
the beam.
Positive internal moment
P
P
-M
-M
Negative internal moment
21
[Reference # 3: Page 569 – 589]
Inflection point
M
22
MomentMoment-curvature relationship
P2
P
ds
P1
y
longitudinal
axis
dx
1
ρ
=−
∈
y
Before deformation
BMD
1
Inflection point
ρ
Elastic curve
Homework # 24: Reference # 3 Problem 12-15
23
=
M
EI
longitudinal
axis
y
o’
ρ ρ
dθ
M
Inflection point
20
Homework # 23: Reference # 3 Problem 6-62
ds’
dx
M
After deformation
24
Slope and displacement by
integration
Represent the curvature
in terms of v and x;
1
From
Simplify by
ρ
1
ρ
=
=
1
1
( )
ρ
ρ
d 2 v / dx 2
[1 + (dv / dx) ]
2 3/ 2
d 2 v / dx 2
M
;
EI
d 2v
dx 2
=
Slope and displacement by
integration
[1 + (dv / dx) ]
2 3/ 2
=
M
EI
d 2v M
=
dx 2 EI
;
From
d 2v M
=
dx 2 EI
From
V=
From
−w=
;
dM
;
dx
dV
;
dx
EI
d 2v
= M ( x)
dx 2
EI
d 3v
= V ( x)
dx 3
EI
d 4v
= − w( x)
dx 4
25
26
Sign convention
Boundary conditions
+w
+M
If positive x is directed to the
left, then θ will be positive
clockwise.
+M
+V
+V
Positive sign convention
v
o’
+ρ
+ρ
ds
+dv
+v
Δ=0
Pin
Δ=0
Roller
θ=0 Δ=0
Fixed end
V=0 M=0
Free end
Δ=0
Pin
Elastic curve
dθ
+θ
x
+x
Δ=0
Roller
dx
M=0
Internal pin or hinge
27
Example
Example
Determine the equations of the elastic curve using the x1 and x2
coordinates. EI is constant.
P
A
a
x2
Determine the maximum deflection of the beam and the slope at
A. EI is constant.
M0
M0
A
B
x1
28
b
B
a
a
a
L
29
Homework # 25: Reference # 3 Problem 12-13
30