Mech 422 Stress and Strain Analysis

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Mech 422 Stress and Strain Analysis

• Course Review

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Plastic Collapse Analysis

• Determine the Plastic Collapse Load for the mild
  steel beam shown below (so36 ksi) using
• (a) Statics
• (b) Energy Methods

Pc
B
C
A
L/2
L/2
L
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Fully Plastic Moment, Mp
Mp soZ Zbh2/4 Mp sobh2/4
b2
bh2/4 2(4)2/4 8 in3
h4
Z
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Plastic Collapse Mechanism
Pc
A
B
C
d
L/2
Hinge,
Mb

• 1 Plastic Hinge is required for the collapse
• Find Pc for Mb Mp

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Statics - Equilibrium - Right Side
Pc
Mp
RcPc/2
L/2
Mp-Rc(L/2)0 but, RcPc/2 MpPcL/4 Pc 4 Mp/L
4sobh2/4L sobh2/L
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Energy Method

• Determine simplest kinematically admissible
  mechanism at collapse

Pc
A
B
a
C
2a
d
L/2
Hinge,
Mb
Assume Elastic deflection is negligible compared
to plastic deflection
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External Work done
This implies Force is constant Hence External
Work, Wp Pd (Recall, for Elastic Deformation
We1/2Pd)
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Geometry Assume d is small
Pc
A
B
a
C
2a
d
L/2
Hinge,
Mb
Tan ad/(L/2) Tan a a (rad, for a small) daL/2
Internal Strain Energy Mp2a Equate External
Work Internal Strain Energy Pd Mp2a
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Pcd Mp2a Pc aL/2 Mp2a
Pc4Mp/L sobh2/L
Same as before.
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Try a harder one
Pc
a
b
B
A
h
C
L
Mp sobh2/4
Find the Plastic Collapse Load, Pc.
Lets use an energy method
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Simplest Collapse Mechanism
Note we need 3 plastic hinges for collapse
Pc
a
A
b
C
B
d
ab q
a
L
a d/a b d/(L-a) ab q d qa(1-a/L)
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Energy Balance

• External Work Pcd
• There are 3 Moment/rotations to store internal
  Strain Energy

U MpaMpbMp(ab) Pcd 2Mp d/ad/(L-a) Pc
2Mp 1/a1/(L-a) Pc 2MpL/a(L-a)
Its that easy!.. ?
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Symmetric caseaL/2
For the symmetric case aL/2
Pc
aL/2
Pc 2MpL/a(L-a) 8Mp/L
L
which is twice the collapse load for the simply
supported beam
Pc
Pc 4 Mp/L
L/2
L/2
L