Beams Session 15-22 Subject: S1014 / MECHANICS of MATERIALS Year: 2008.
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Transcript of Beams Session 15-22 Subject: S1014 / MECHANICS of MATERIALS Year: 2008.
Beams Session 15-22
Subject : S1014 / MECHANICS of MATERIALSYear : 2008
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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 StressLike 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 ρ
= – ε y
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ELASTIC CURVE
Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/IWe have
1 ρ = M
EI or
1 ρ
= – σ 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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This assumes that the system is linear-elastic, and therefore the deflection is a linear function of F.
, Deflection at B
Loa
d, F
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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.
ii
ii
i AE
LPU
2
27
1
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Beam Elements:
y
x
y
dxdA
z
y
F(x)
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:
y
x
y
dxdA
z
y
F(x)
The Strain Energy Density on this element is:
2
2
12
1
x
xx
E
u
For linear elastic material
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Substituting,
I
Myx dxdA
EI
yMudxdA
2
22
2
and multiplying by the Volume of the element
Hence, the Strain Energy for a slice of the beam, of width dx, is
A
A
dAydxEI
M
udxdAdU
22
2
2
y
x
dx
xxAIdAy 2
dxEI
MdU
2
2
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Strain Energy in Entire Beam
Consider the cantilever beam as shown
I
FL
x
y
M=F(x-L)
dxEI
LxFU
L2
0 2
EI
LF
6
32
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Deflection
I
L
x
y
F
FW2
1 =
EI
LF
6
32
EI
FL
3
3
External Work, Strain Energy
Linear-elastic,F
Classical Solution
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y
x
P
L/2 L/2
She
ar F
orce
PL/4
P/2
-P/2M
omen
t
Determine Elastic Strain Energy due to bending for simply supported 3-point bending member of constant X-section.
For 0 xL/2: M=Px/2
Note by symmetry we can find the total strain energy by doubling the strain energy of the LHS.
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EI
LP
EI
xP
dxEI
xP
dxEI
Mdx
EI
MU
L
L
LL
96
12
82
22
2
32
2/
0
32
2/
0
22
2/
0
2
0
2
y
L/2 L/2
P
B
Determine B…….
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Elastic Strain Energy due to Transverse Shear Stress
y
x
3
2
1aU yxy
xya
xy
G
Gu
xy
xyxyxyxy
2
;2
1
2
xy
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Shear Strain Energy
y
dA
z
y
x
dx
F(x)
dxAG
fT
dAdxGA
TU
forceshearTWheredAT
dVG
udVU
xy
xy
2
2
2
2
2
1
2
1
;/2
1
f is called a form factor:Circle f=1.11Rectangle f=1.2Tube f=2.00I section f=A/Aweb
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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• Stiffness• Stability – Important in compressive members
• Stress and strain relationships can be studied with Mohr’s circle
Often the controlling factor for functionality
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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 insufficientWe can use energy-methods in these circumstancesDefine strain energy
1
0
x
FdxU
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Deflection --- Energy Method
• Define strain energy density**V – volume
dV
dU
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Deflection --- Energy Method
Put in terms of ,
dVE
U
dUdVdV
dUE
E
x
xxx
xx
2
2
2
1
2
1
2
1
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Example – beam in bending
)(2
2
2
2
2
2
22
2
xfEI
M
dAdxdV
dVEI
yMU
dVE
U
I
My
x
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Example – beam in bending
dxEI
MU
dxEI
dAyMdAdx
EI
yMdV
EI
yMU
dAyI
2
2)(
222
2
22
2
22
2
22
2
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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
ii F
U
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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
• 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
Materialfailure
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Unit Load Method
• Deflection at C ???
A BC
L
q
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Unit Load Method
• Procedure 1 : Determine Mo
A BC
L
q
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Unit Load Method
Procedure 1 : Determine Mo
q
VA
HA
VB
MA=0w. ½ L- VB.L = 0qL ½ L- VB.L= 0½ qL2 – VB.L = 0
VB = ½ qL
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Unit Load Method
Procedure 1 : Determine Mo
q
VA
HA
VB
VA = ½ qL
MB=0- w. ½ L+ VA.L = 0- qL ½ L+ VA.L= 0- ½ qL2 + VA.L = 0
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Unit Load Method
Procedure 1 : Determine Mo
q
VA
HA
VB
H=0HA = 0
HA = 0
V=0
VA + VB = w½ qL + ½ qL = qL
OK!!!
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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for A-C 0<x1< 1/2 L
HA
VA
q
A
w = qx1
x1
Mx=0w.( ½ .x1) –VA.x1 + Mx1 = 0( qx1.x1 ) .( ½ .x1) – qL.x1 + Mx1 = 0Mx1 = qLx1 – ½ x1
2
Mx1 = qLx1 – ½ x12
Mx1
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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for B-C 0<x2< 1/2 L
Mx=0w.( ½ .x2) –VB.x2 + Mx2 = 0( qx1.x2 ) .( ½ .x2) – qL.x2 + Mx2 = 0Mx2 = qLx2 – ½ x2
2
Mx2 = qLx2 – ½ x22
VB
B
x2 q
w = qx2
Mx2
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Unit Load MethodProcedure 2 : Put P = 1 unit on C ( without external loads )
VA
HA
VB
MA=0P. ½ L- VB.L = 01 ½ L- VB.L= 0½ L – VB.L = 0
VB = ½
P = 1 unit
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Unit Load MethodProcedure 2 : Put P = 1 unit on C ( without external loads )
VA
HA
VB
VA = ½
MB=0- P. ½ L+ VA.L = 0- 1. ½ L+ VA.L= 0- ½ L + VA.L = 0
P = 1 unit
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Unit Load MethodProcedure 2 : Put P = 1 unit on C ( without external loads )
VA
HA
VB
H=0HA = 0
HA = 0
V=0
VA + VB = w½ + ½ = 1 OK!!!
P = 1 unit
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Unit Load Method
Procedure 3 : Determine m for A-C 0<x1< 1/2 L
HA
VA
Ax1
Mx=0–VA.x1 + mx1 = 0– ½ .x1 + mx1 = 0mx1 = ½ x1
mx1 = ½ x1
mx1
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Unit Load MethodProcedure 3 : Determine m ( cont’ ) for B-C 0<x2< 1/2 L
Mx=0–VB.x2 + Mx2 = 0– ½ .x2 + Mx2 = 0Mx2 = ½ x2
Mx2 = ½ x2
VB
B
x2 q
Mx2
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Unit Load MethodProcedure 4 : Determine deflection at C
Mx1 = qLx1 – ½ x12 Mx2 = qLx2 – ½ x2
2
mx1 = ½ x1 Mx2 = ½ x2
A-C0<x1< ½ L
B-C0<x2< ½ L
odxEI
mxMx
.
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Unit Load Method
Procedure 4 : Determine deflection at C
2
2/1
0
2221
2/1
0
111 )(1/2x*)2^ x½– (qLx)(1/2x*)2^ x½– (qLxdx
EIdx
EI
LL
dxEI
mxMx
.
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Unit Load Method
Procedure 4 : Determine deflection at C
384
5
EI
qL
2
2/1
0
2221
2/1
0
111 )(1/2x*)2^ x½– (qLx)(1/2x*)2^ x½– (qLxdx
EIdx
EI
LL
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Unit Load Method
Deflection at C =
A BC
L
q 384
5
EI
qL
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Unit Load Method
• Slope Deflection C ???
A BC
L
q
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Unit Load Method
• Procedure 1 : Determine Mo
A BC
L
q
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Unit Load Method
Procedure 1 : Determine Mo
q
VA
HA
VB
MA=0w. ½ L- VB.L = 0qL ½ L- VB.L= 0½ qL2 – VB.L = 0
VB = ½ qL
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Unit Load Method
Procedure 1 : Determine Mo
q
VA
HA
VB
VA = ½ qL
MB=0- w. ½ L+ VA.L = 0- qL ½ L+ VA.L= 0- ½ qL2 + VA.L = 0
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Unit Load Method
Procedure 1 : Determine Mo
q
VA
HA
VB
H=0HA = 0
HA = 0
V=0
VA + VB = w½ qL + ½ qL = qL
OK!!!
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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for A-C 0<x1< 1/2 L
HA
VA
q
A
w = qx1
x1
Mx=0w.( ½ .x1) –VA.x1 + Mx1 = 0( qx1.x1 ) .( ½ .x1) – qL.x1 + Mx1 = 0Mx1 = qLx1 – ½ x1
2
Mx1 = qLx1 – ½ x12
Mx1
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Unit Load MethodProcedure 1 : Determine Mo ( cont’ ) for B-C 0<x2< 1/2 L
Mx=0w.( ½ .x2) –VB.x2 + Mx2 = 0( qx1.x2 ) .( ½ .x2) – qL.x2 + Mx2 = 0Mx2 = qLx2 – ½ x2
2
Mx2 = qLx2 – ½ x22
VB
B
x2 q
w = qx2
Mx2
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Unit Load MethodProcedure 2 : Put M = 1 unit on C ( without external loads )
MA=0M - VB.L = 01 - VB.L= 0
VB = 1/L
VA
HA
VB
M = 1 unit
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Unit Load MethodProcedure 2 : Put M = 1 unit on C ( without external loads )
MB=0M + VA.L = 01 + VA.L= 0
VA = -1/L
VA
HA
VB
M = 1 unit
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Unit Load MethodProcedure 2 : Put M = 1 unit on C ( without external loads )
VA
HA
VB
M = 1 unit
H=0HA = 0
HA = 0
V=0
-VA + VB = 0- ½L + ½L = 0 OK!!!
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Unit Load Method
Procedure 3 : Determine m for A-C 0<x1< 1/2 L
HA
VA
Ax1
Mx=0–VA.x1 + mx1 = 0– -1/L .x1 + mx1 = 0mx1 = -1/L x1
mx1 = -1/L x1
mx1
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Unit Load MethodProcedure 3 : Determine m ( cont’ ) for B-C 0<x2< 1/2 L
Mx=0–VB.x2 + Mx2 = 0– 1/L.x2 + Mx2 = 0Mx2 = 1/Lx2
Mx2 = 1/L x2
VB
B
x2 q
Mx2
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Unit Load MethodProcedure 4 : Determine deflection at C
Mx1 = qLx1 – ½ x12 Mx2 = qLx2 – ½ x2
2
mx1 = -1/Lx1 Mx2 = 1/Lx2
A-C0<x1< ½ L
B-C0<x2< ½ L
o
dxEI
mxMx
.
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Unit Load Method
Procedure 4 : Determine deflection at C
2
2/1
0
2221
2/1
0
111 )(1/Lx*)2^ x½– (qLx)(-1/Lx*)2^ x½– (qLxdx
EIdx
EI
L
o
L
o
dxEI
mxMx
.
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Unit Load Method
Procedure 4 : Determine deflection at C
0
2
2/1
0
2221
2/1
0
111 )(1/Lx*)2^ x½– (qLx)(-1/Lx*)2^ x½– (qLxdx
EIdx
EI
L
o
L
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Unit Load Method
Slope Deflection C = 0
A BC
L
q