Subject
Year
: S1014 / MECHANICS of MATERIALS
: 2008
Beams
Session 15-22
Beams
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What is Bending Stresses ?
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What is Bending Stresses ?
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Normal Stress
A normal stress is a stress that occurs when
a member is loaded by an axial force.
The value of the normal force for any
prismatic section is simply the force divided
by the cross sectional area.
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Normal Stress
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What is Bending Stresses ?
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What is Bending Stresses ?
When a member is being loaded similar
to that in figure 1 bending stress (or
flexure stress) will result. Bending
stress is a more specific type of normal
stress.
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What is Bending Stresses ?
When a beam experiences load like that
shown in figure 1 the top fibers of the
beam undergo a normal
compressive stress.
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What is Bending Stresses ?
The stress at the horizontal plane of the
neutral is zero. The bottom fibers of
the beam undergo a normal tensile
stress.
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What is Bending Stresses ?
It can be concluded therefore
that the value of the bending
stress will vary linearly with
distance from the neutral axis.
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What is Bending Stresses ?
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Shear Stress
Normal stress is a result of load applied
perpendicular to a member.
Shear stress however results when a load is
applied parallel to an area.
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Shear Stress
Like in bending stress, shear
stress will vary across the
cross sectional area.
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Shear Stress
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ELASTIC CURVE
The deflection diagram of the longitudinal axis that passes through the centroid of
each cross-sectional area of the beam is called the elastic curve, which is
characterized by the deflection and slope along the curve. E.g.
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ELASTIC CURVE
Moment-curvature relationship:
Sign convention:
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ELASTIC CURVE
Moment-curvature relationship:
Sign convention:
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ELASTIC CURVE
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ELASTIC CURVE
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ELASTIC CURVE
Consider a segment of width dx, the strain in are ds, located at a
position y from the neutral axis is ε = (ds’ – ds)/ds.
However, ds = dx = ρdθ and ds’ = (ρ-y) dθ,
and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or
1 =– ε
ρ
y
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ELASTIC CURVE
Comparing with the Hooke’s Law ε = σ / E and the flexure
formula σ = -My/I
We have
1
M
=
ρ
EI
or
1 =– σ
ρ
Ey
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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)
Boundary Conditions:
• The integration constants can be
determined by imposing the boundary
conditions, or
• Continuity condition at specific
locations
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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)
Boundary Conditions:
• The integration constants can be
determined by imposing the boundary
conditions, or
• Continuity condition at specific
locations
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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)
Boundary Conditions:
• The integration constants can be
determined by imposing the boundary
conditions, or
• Continuity condition at specific
locations
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SLOPE AND DISPLACEMENT BY INTEGATION (CONT.)
Boundary Conditions:
• The integration constants can be
determined by imposing the boundary
conditions, or
• Continuity condition at specific
locations
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Load, F
This assumes that the system is linear-elastic, and
therefore the deflection D is a linear function of F.
D, Deflection at B
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The total strain energy stored in the system is the sum of the
individual strain energies in each of the truss members numbered
i=1 to 7.
7
U 
i 1
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2
Pi Li
2 Ei Ai
Beam Elements:
F(x)
x
z
y
dA
dx
y
y
A beam that is symmetrical in x-section about the z-axis, is
subjected to bending. Consider a infinitesimal volume element of
length dx and area dA as shown. This element is subjected to a
normal stress: sx=My/I
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Beam Elements:
F(x)
x
z
y
dA
dx
y
y
The Strain Energy Density on this element is:
For linear elastic material
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1
u   x x
2
1 2

x
2E
Substituting,
and multiplying by the Volume of the element
My
x 
I
M 2 y2
udxdA 
dxdA
2
2 EI
Hence, the Strain Energy for a slice of the beam, of width dx, is
dU   udxdA
A
x

A
dx
y
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M2
2

dx
y
dA
2

A
2 EI
y 2 dA  I xx
M2
dU 
dx
2 EI
Strain Energy in Entire Beam
Consider the cantilever beam as shown
F
L
x
I
y
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d
M=F(x-L)
U 

L
0
F x  L 2 dx
F 2 L3

6 EI
2 EI
Deflection
L
F
x
I
y
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External Work,
Linear-elastic,
Fd
d
1
W  Fd =
2
2 3
F L

6 EI
3
FL
d
3EI
Strain Energy
Classical Solution
Note by symmetry we
can find the total strain
energy by doubling the
strain energy of the LHS.
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L/2
L/2
y
Shear Force
For 0  xL/2: M=Px/2
x
P/2
-P/2
PL/4
Moment
Determine Elastic Strain
Energy due to bending for
simply supported 3-point
bending member of
constant X-section.
P
L
L/2
M
M2
U 
dx  2 
dx
2 EI
2 EI
0
0
L/2
2
2
0
P x
dx
8 EI
2
3 L/2
2
P x

12 EI
P 2 L3

96 EI
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0
P
2
L/2
DB
y
Determine DB…….
L/2
Elastic Strain Energy due to Transverse Shear Stress
y
xy
1
U   xy y a 3
2
1
u   xy xy ;  xy  G xy
2
xy
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x
d  xya

 xy 2
2G
Shear Strain Energy
F(x)
x
z
y
dA
dx
y
1  xy
U   udV  
dV
2 G
 xy  T / dA;Where T  shear force
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1 T2
U   2 dAdx
2 AG
1 fT 2
 
dx
2 AG
2
f is called a form factor:
Circle
f=1.11
Rectangle f=1.2
Tube
f=2.00
I section f=A/Aweb
Applications:
• Castigliano’s 2nd theorem can be used to determine the
deflections in structures (eg, trusses, beams, frames,
shells) and we are not limited to applications in which
only 1 external force or moment acts.
• Furthermore, we can determine the deflection or
rotation at any point, even where no force or moment
is applied externally.
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Design Considerations
•
•
•
•
•
Stress – Yield Failure or Code Compliance
Deflection
Strain
Often the controlling factor for
Stiffness
functionality
Stability – Important in compressive members
• Stress and strain relationships can be studied with Mohr’s circle
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Deflection [Everything’s a Spring]
• When loads are applied, we have deflection
• Depends on
– Type of loading
• Tension
• Compression
• Bending
• Torsion
– Cross-section of member
– Comparable to pushing on a spring
• We can calculate the amount of beam deflection by various
methods
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Superposition
• Determine effects of individual loads separately and add
the results
• May be applied if
– Each effect is linearly related to the load that produces it
– A load does not create a condition that affects the result of
another load
– Deformations resulting from any specific load are not large
enough to appreciably alter the geometric relations of the parts
of the structural system
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Deflection --- Energy Method
There are situations where the tables are
insufficient
We can use energy-methods in these circumstances
Define strain energy
U 
x1
Fdx

0
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Deflection --- Energy Method
• Define strain energy density**
V – volume
dU

dV
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Deflection --- Energy Method
Put in terms of , 
 x  E x
1
1 x
   x x 
2
2 E
dU

dV
dV  dU
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 1  x2 
dV
U   

2
E


2
Example – beam in bending
My
 
I
U 
x
2
 2 E dV
M 2 y2
U 
dV
2
2 EI
dV  dAdx
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M2
 f ( x)
2
2 EI
Example – beam in bending
I   y 2 dA
2
2
2
2
M y
M y
U 
dV  
( dAdx )  
2
2
2 EI
2 EI
M2
U 
dx
2 EI
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M2
 y dAdx
2
2 EI 2
Castigliano’s Theorem
• Deflection at any point along a beam subjected to n loads may be
expressed as the partial derivative of the strain energy of the
structure WRT the load at that point
U
di 
Fi
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Castigliano’s Theorem
• We can derive the strain energy equations as we did for bending
• Then we take the partial derivative to determine the deflection
equation
• AND if we don’t have a force at the desired point:
– If there is no load acting at the point of interest, add a dummy load Q, work
out equations, then set Q = 0
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Stability
• Up until now, 2 primary concerns
– Strength of a structure
Material
failure
• It’s ability to support a specified load without experiencing excessive
stress
– Ability of a structure to support a specified load without undergoing
unacceptable deformations
• Now, look at STABILITY of the structure
– It’s ability to support a load without undergoing a sudden change in
configuration
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Unit Load Method
• Deflection at C ???
A
q
C
L
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B
Unit Load Method
• Procedure 1 : Determine Mo
A
q
C
L
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B
Unit Load Method
Procedure 1 : Determine Mo
q
HA
VA
VB
SMA=0
w. ½ L- VB.L = 0
qL ½ L- VB.L= 0
½ qL2 – VB.L = 0
Bina Nusantara
VB = ½ qL
Unit Load Method
Procedure 1 : Determine Mo
q
HA
VA
VB
SMB=0
- w. ½ L+ VA.L = 0
- qL ½ L+ VA.L= 0
- ½ qL2 + VA.L = 0
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VA = ½ qL
Unit Load Method
Procedure 1 : Determine Mo
q
HA
VA
VB
SH=0
HA = 0
HA = 0
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SV=0
VA + VB = w
½ qL + ½ qL = qL
OK!!!
Unit Load Method
Procedure 1 : Determine Mo ( cont’ ) for A-C  0<x1< 1/2 L
x1
q
Mx1
A
HA
w = qx1
VA
SMx=0
w.( ½ .x1) –VA.x1 + Mx1 = 0
( qx1.x1 ) .( ½ .x1) – qL.x1 + Mx1 = 0
Mx1 = qLx1 – ½ x12
Mx1 = qLx1 – ½ x12
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Unit Load Method
Procedure 1 : Determine Mo ( cont’ ) for B-C  0<x2< 1/2 L
x2
q
B
Mx2
w = qx2
VB
SMx=0
w.( ½ .x2) –VB.x2 + Mx2 = 0
( qx1.x2 ) .( ½ .x2) – qL.x2 + Mx2 = 0
Mx2 = qLx2 – ½ x22
Mx2 = qLx2 – ½ x22
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Unit Load Method
Procedure 2 : Put P = 1 unit on C ( without external loads )
P = 1 unit
HA
VA
VB
SMA=0
P. ½ L- VB.L = 0
1 ½ L- VB.L= 0
½ L – VB.L = 0
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VB = ½
Unit Load Method
Procedure 2 : Put P = 1 unit on C ( without external loads )
P = 1 unit
HA
VA
VB
SMB=0
- P. ½ L+ VA.L = 0
- 1. ½ L+ VA.L= 0
- ½ L + VA.L = 0
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VA = ½
Unit Load Method
Procedure 2 : Put P = 1 unit on C ( without external loads )
P = 1 unit
HA
VA
VB
SH=0
HA = 0
SV=0
VA + VB = w
½ +½ =1
HA = 0
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OK!!!
Unit Load Method
Procedure 3 : Determine m for A-C  0<x1< 1/2 L
x1
mx1
A
HA
VA
SMx=0
–VA.x1 + mx1 = 0
– ½ .x1 + mx1 = 0
mx1 = ½ x1
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mx1 = ½ x1
Unit Load Method
Procedure 3 : Determine m ( cont’ ) for B-C  0<x2< 1/2 L
x2
q
Mx2
B
VB
SMx=0
–VB.x2 + Mx2 = 0
– ½ .x2 + Mx2 = 0
Mx2 = ½ x2
Bina Nusantara
Mx2 = ½ x2
Unit Load Method
Procedure 4 : Determine deflection at C
A-C
0<x1< ½ L
Mx1 = qLx1 – ½ x12
mx1 = ½ x1
B-C
0<x2< ½ L
Mx2 = qLx2 – ½ x22
Mx2 = ½ x2
Mx.mx
d 
dxo
EI
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Unit Load Method
Procedure 4 : Determine deflection at C
Mx.mx
d 
dx
EI
d
1/ 2 L

0
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(qLx 1 – ½ x1^ 2) * (1/2x 1)
dx1 
EI
1/ 2 L

0
(qLx2 – ½ x2^ 2) * (1/2x 2)
dx 2
EI
Unit Load Method
Procedure 4 : Determine deflection at C
d
1/ 2 L

0
(qLx 1 – ½ x1^ 2) * (1/2x 1)
dx1 
EI
1/ 2 L

0
(qLx2 – ½ x2^ 2) * (1/2x 2)
dx 2
EI
5 qL
d
384 EI
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Unit Load Method
Deflection at C =
A
5 qL
d
384 EI
C
L
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q
B
Unit Load Method
• Slope Deflection C ???
A
q
C
L
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B
Unit Load Method
• Procedure 1 : Determine Mo
A
q
C
L
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B
Unit Load Method
Procedure 1 : Determine Mo
q
HA
VA
VB
SMA=0
w. ½ L- VB.L = 0
qL ½ L- VB.L= 0
½ qL2 – VB.L = 0
Bina Nusantara
VB = ½ qL
Unit Load Method
Procedure 1 : Determine Mo
q
HA
VA
VB
SMB=0
- w. ½ L+ VA.L = 0
- qL ½ L+ VA.L= 0
- ½ qL2 + VA.L = 0
Bina Nusantara
VA = ½ qL
Unit Load Method
Procedure 1 : Determine Mo
q
HA
VA
VB
SH=0
HA = 0
HA = 0
Bina Nusantara
SV=0
VA + VB = w
½ qL + ½ qL = qL
OK!!!
Unit Load Method
Procedure 1 : Determine Mo ( cont’ ) for A-C  0<x1< 1/2 L
x1
q
Mx1
A
HA
w = qx1
VA
SMx=0
w.( ½ .x1) –VA.x1 + Mx1 = 0
( qx1.x1 ) .( ½ .x1) – qL.x1 + Mx1 = 0
Mx1 = qLx1 – ½ x12
Mx1 = qLx1 – ½ x12
Bina Nusantara
Unit Load Method
Procedure 1 : Determine Mo ( cont’ ) for B-C  0<x2< 1/2 L
x2
q
B
Mx2
w = qx2
VB
SMx=0
w.( ½ .x2) –VB.x2 + Mx2 = 0
( qx1.x2 ) .( ½ .x2) – qL.x2 + Mx2 = 0
Mx2 = qLx2 – ½ x22
Mx2 = qLx2 – ½ x22
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Unit Load Method
Procedure 2 : Put M = 1 unit on C ( without external loads )
M = 1 unit
HA
VA
VB
SMA=0
M - VB.L = 0
1 - VB.L= 0
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VB = 1/L
Unit Load Method
Procedure 2 : Put M = 1 unit on C ( without external loads )
M = 1 unit
HA
VA
VB
SMB=0
M + VA.L = 0
1 + VA.L= 0
Bina Nusantara
VA = -1/L
Unit Load Method
Procedure 2 : Put M = 1 unit on C ( without external loads )
M = 1 unit
HA
VA
VB
SH=0
HA = 0
SV=0
-VA + VB = 0
- ½L + ½L = 0
HA = 0
Bina Nusantara
OK!!!
Unit Load Method
Procedure 3 : Determine m for A-C  0<x1< 1/2 L
x1
mx1
A
HA
VA
SMx=0
–VA.x1 + mx1 = 0
– -1/L .x1 + mx1 = 0
mx1 = -1/L x1
Bina Nusantara
mx1 = -1/L x1
Unit Load Method
Procedure 3 : Determine m ( cont’ ) for B-C  0<x2< 1/2 L
x2
q
Mx2
B
VB
SMx=0
–VB.x2 + Mx2 = 0
– 1/L.x2 + Mx2 = 0
Mx2 = 1/Lx2
Bina Nusantara
Mx2 = 1/L x2
Unit Load Method
Procedure 4 : Determine deflection at C
A-C
0<x1< ½ L
Mx1 = qLx1 – ½ x12
mx1 = -1/Lx1
B-C
0<x2< ½ L
Mx2 = qLx2 – ½ x22
Mx2 = 1/Lx2
Mx.mx
 
dx
EI
o
Bina Nusantara
Unit Load Method
Procedure 4 : Determine deflection at C
Mx.mx
 
dx
EI
o

1/ 2 L

0
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(qLx 1 – ½ x1^ 2) * (-1/Lx1)
dx1 
EI
o
1/ 2 L

0
(qLx 2 – ½ x2^ 2) * (1/Lx 2)
dx 2
EI
Unit Load Method
Procedure 4 : Determine deflection at C

1/ 2 L

0
(qLx 1 – ½ x1^ 2) * (-1/Lx1)
dx1 
EI
o
 0
Bina Nusantara
1/ 2 L

0
(qLx 2 – ½ x2^ 2) * (1/Lx 2)
dx 2
EI
Unit Load Method
Slope Deflection C = 0
A
q
C
L
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B