Part (1) basics of plastic analysis

PLASTIC ANALYSIS AND DESIGN OF
STEEL STRUCTURES
CES 608
Port Said University
Faculty of Engineering
Civil Engineering Department
BASICS OF PLASTIC DESIGN
2019

PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES
Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt
L E C T U R E 1 : B A S I C S O F P L A S T I C D E S I G N 2
1. Introduction
• In the elastic analysis of structures superposition is
often used and it is valid.
• However, an elastic analysis does not give information
about the loads that will collapse a structure.
• An indeterminate structure may sustain loads greater
than the load that first causes a yield to occur at any
point in the structure.

• In fact, a structure will stand if it is able to find
redundancies to yield.
• It is only when a structure has exhausted all its
redundancies will extra load causes it to fail.
• Plastic analysis is the method through which the actual
failure load of a structure is calculated, and as will be
seen, this failure load can be significantly greater than
the elastic load capacity.
1. Introduction

𝝈
𝜺
𝟏
𝑬
Upper yield
Lower yield
Necking
Failure
Elastic
Zone
Plastic
Zone
Strain
Hardening
The material can sustain strains far in excess of the strain at which
yield occurs before failure.
This property of the material is called its ductility.
2. Material Behavior

For plastic analysis, we change the typical diagram by its
idealized diagram. Different idealized diagrams are
considered in engineering practice. Some of idealized
models are presented:
Ref.: Igor A. Karnovsky and Olga Lebed, “Advanced Methods of Structural Analysis”

𝝈
𝜺
𝟏
𝑬
Yield point
Necking
Failure
Elastic
Zone
Plastic
Zone
Strain
Hardening
𝝈 𝒚
𝝈 𝒖
𝜺 𝒚 𝜺 𝒔𝒉 𝜺 𝒖

𝝈
𝜺
𝟏
𝑬
Yield point
Elastic
Zone
Plastic
Zone
complex models exist to represent the real behavior of the material.
The most common, and simplest, model is the idealized stress-strain
curve (for an ideal elastic-plastic material - which doesn’t exist).
𝝈 𝒚
𝜺 𝒚 𝜺 𝒖
Failure
Perfectly Plastic Material Curve
Ideal Elastic-Plastic Material Curve

𝝈
𝜺
𝟏
𝑬
𝝈 𝒚
𝜺 𝒚
Stress-Strain
Curve for Elastic-
Perfectly Plastic
Material
Tension
Compression
− 𝝈 𝒚
−𝜺 𝒚
Horizontal Plateau

• Since so much post-yield strain is modelled, the actual
material (or cross section) must also be capable of
allowing such strains.
• That is, it must be sufficiently ductile for the idealized
stress-strain curve to be valid.
• So, we consider the behavior of a cross section of an
ideal elastic-plastic material subject to bending.
• In doing so, we seek the relationship between applied
moment and the rotation (or more accurately, the
curvature) of a cross section.
𝝈
𝜺
𝟏
𝑬
𝝈 𝒚
𝜺 𝒚
• Once the yield has been
reached it is taken that an
indefinite amount of strain can
occur.
3. Behavior of Cross-Section

• Consider an arbitrary cross-section with a vertical
plane of symmetry, which is also the plane of loading.
• Consider the cross section subject to an increasing
bending moment and assess the stresses at each
stage.
𝐸. 𝑁. 𝐴
𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠
𝑧
𝑦
𝐶. 𝐺

𝐸. 𝑁. 𝐴
𝑃. 𝑁. 𝐴
𝐸. 𝑁. 𝐴
𝑃. 𝑁. 𝐴
𝜀 < 𝜀 𝑦
𝜀 < 𝜀 𝑦 Strain
𝜎 < 𝜎 𝑦
𝜎 < 𝜎 𝑦
𝜀 = 𝜀 𝑦
𝜀 < 𝜀 𝑦
𝜎 = 𝜎 𝑦
𝜎 < 𝜎 𝑦
𝜀 > 𝜀 𝑦
𝜀 = 𝜀 𝑦
𝜎 = 𝜎 𝑦
𝜎 = 𝜎 𝑦
𝜀 > 𝜀 𝑦
𝜀 > 𝜀 𝑦
𝜎 = 𝜎 𝑦
𝜎 = 𝜎 𝑦
𝜀 > 𝜀 𝑦
𝜀 > 𝜀 𝑦
𝜎 = 𝜎 𝑦
𝜎 = 𝜎 𝑦 Stress
Elastic Elasto-Plastic Fully Plastic
𝛼=1
𝛼<1
𝛼=0
𝛼<1
α = the ratio of the depth of the elastic to plastic regions.
𝜎 = 𝜎 𝑦
𝛼=0
Strain hardening

𝑹𝒐𝒕𝒂𝒕𝒊𝒐𝒏
Failure
Elastic
Elasto-
Plastic
Strain
Hardening
𝑴 𝒚
𝑴 𝒑
𝑴
Plastic
𝜎 = 𝜎 𝑦
𝜎 = 𝜎 𝑦
𝛼<1
𝜎 = 𝜎 𝑦
𝜎 < 𝜎 𝑦
𝛼=1
𝜎 = 𝜎 𝑦
𝜎 = 𝜎 𝑦
𝛼=0 𝜎 = 𝜎
𝜎 = 𝜎 𝑦
𝛼=0
𝜀 > 𝜀 𝑦
𝜀 = 𝜀 𝑦
𝜀 = 𝜀
𝜀 𝑦 < 𝜀 < 𝜀

𝑹𝒐𝒕𝒂𝒕𝒊𝒐𝒏
Failure
Elastic
Elasto-
Plastic
𝑴 𝒚
𝑴 𝒑
𝑴
Plastic
𝜎 = 𝜎 𝑦
𝜎 = 𝜎 𝑦
𝛼<1
𝜎 = 𝜎 𝑦
𝜎 < 𝜎 𝑦
𝛼=1
𝜎 = 𝜎 𝑦
𝜎 = 𝜎 𝑦
𝛼=0
𝜀 > 𝜀 𝑦
𝜀 = 𝜀 𝑦
𝜀 = 𝜀
𝜀 𝑦 < 𝜀 < 𝜀
Yield point
Elastic
Zone
Plastic
Zone
𝝈 𝒚
𝜺 𝒚 𝜺 𝒖
Failure

Plastic Hinge
• With idealized moment-rotation curve, the cross section linearly
sustains moment up to the plastic moment capacity of the section
and then yields in rotation an indeterminate amount.
• Again, to use this idealization, the actual section must be capable of
sustaining large rotations – that is it must be ductile.
• Once the plastic moment capacity is reached, the section can
rotate freely (that is, it behaves like a hinge, except with moment of
Mp at the hinge).
• This is termed a plastic hinge and it is the basis for plastic analysis.
• At the plastic hinge stresses remain constant, but strains and hence
rotations can increase.

Assumptions:
• The material obeys Hooke's law until the stress reaches the yield
value and thereafter remains constant.
• The yield stresses and the modulus of elasticity have the same value
in compression as in tension.
• The material is homogeneous and isotropic in both the elastic and
plastic states.
• The plane transverse sections (the sections perpendicular to the
longitudinal axis of the beam) remain plane and normal to the
longitudinal axis after bending, the effect of shear being neglected.
• There is no resultant axial force on the beam.
• The cross section of the beam is symmetrical about an axis through
its centroid parallel to plane of bending.
• Every layer of the material is free to expand and contract
longitudinally and laterally under the stress as if separated from the
other layers.
4. Analysis of Cross-Section

𝐸. 𝑁. 𝐴
𝑃. 𝑁. 𝐴
𝜎 = 𝜎 𝑦
ത𝑦𝑐
𝐶
𝜎 = 𝜎 𝑦𝜎 ≤ 𝜎 𝑦
𝜎 = 𝜎 𝑦
𝐶. 𝐺
𝑃𝑙𝑎𝑠𝑡𝑖𝑐
𝐶𝑒𝑛𝑡𝑒𝑟
𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪
𝑇
𝐶
𝑇
𝑦𝑡
𝑦 𝑏
ത𝑦𝑡
𝐴 𝑐
𝐴 𝑡
𝐶 = 𝑇
𝑀 𝑦 = 𝜎 𝑦 ×
𝐼
𝑦𝑡
𝑀 𝑦 = 𝜎 𝑦 × 𝑍 𝑒
𝑀 𝑦 = yield moment
𝑍 𝑒 = elastic section modulus
𝐶 = 𝑇 𝜎 𝑦 𝐴 𝑐 = 𝜎 𝑦 𝐴 𝑡
𝐴 𝑐 = 𝐴 𝑡 = 𝐴/2
𝑀 𝑝 = 𝜎 𝑦 (𝐴 𝑐 ത𝑦𝑐 + 𝐴 𝑡 ത𝑦𝑡) =
𝜎 𝑦 𝐴
2
(ത𝑦𝑐 + ത𝑦𝑡)
𝑀 𝑝 = 𝜎 𝑦 × 𝑍 𝑝
→
𝑀 𝑝 = plastic moment
𝑍 𝑝 = plastic section modulus
4. Analysis of Cross-Sections

𝐸. 𝑁. 𝐴
𝜎 𝑦
𝐶. 𝐺
𝑬𝑳𝑨𝑺𝑻𝑰𝑪
𝑇
𝑑
𝑏
2
3
𝑑 𝛼𝑑
2
3
𝛼𝑑
ൗ𝑑
2
𝐶 𝐶1
𝐶2
𝑇1
𝑇2
𝑇
𝐶
𝜎 𝑦
𝜎 𝑦 𝜎 𝑦
𝜎 𝑦 𝜎 𝑦𝑑 1 − 𝛼
4
𝑑 1 + 𝛼
2
𝑃. 𝑁. 𝐴
𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪
ൗ𝑑
2
ൗ𝑑
2
𝑍 𝑒 = 𝑏 𝑑2
/6
𝑀 𝑦 = 𝜎 𝑦 𝑍 𝑒
𝑀 𝑦 = 𝜎 𝑦
𝑏 𝑑2
6
𝐶 = 𝑇 =
1
2
𝜎 𝑦
𝑑
2
𝑏 = 𝜎 𝑦
𝑏 𝑑
4
𝑀 𝑦 = 𝐶 ×
2
3
𝑑 = 𝜎 𝑦
𝑏 𝑑
4
×
2
3
𝑑
𝑀 𝑦 = 𝜎 𝑦
𝑏 𝑑2
6
Alternative method using stress block:
M
4.1. Rectangular Cross-Section

𝐸. 𝑁. 𝐴
𝜎 𝑦
𝐶. 𝐺
𝑇
𝑑
𝑏
2
3
𝑑 𝛼𝑑
2
3
𝛼𝑑
ൗ𝑑
2
𝐶 𝐶1
𝐶2
𝑇1
𝑇2
𝑇
𝐶
𝜎 𝑦
𝜎 𝑦 𝜎 𝑦
4
𝑑 1 + 𝛼
2
𝑃. 𝑁. 𝐴
ൗ𝑑
2
ൗ𝑑
2
𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪
𝐶1 = 𝑇1 = 𝜎 𝑦
𝑑 (1 − 𝛼)
2
𝑏 , 𝐴𝑟𝑚1 =
𝑑 (1 + 𝛼)
2
𝑀𝑒𝑝 = 𝐶1 × 𝐴𝑟𝑚1 + 𝐶2 × 𝐴𝑟𝑚2
𝑀𝑒𝑝 = 𝜎 𝑦
𝑑 (1 − 𝛼)
2
𝑏 ×
𝑑 (1 + 𝛼)
2
+
1
2
𝜎 𝑦
𝛼 𝑑
2
𝑏 ×
2
3
𝛼 𝑑
using stress block:
𝐶2 = 𝑇2 =
1
2
𝜎 𝑦
𝛼 𝑑
2
𝑏 , 𝐴𝑟𝑚2 =
2
3
𝛼 𝑑

𝐸. 𝑁. 𝐴
𝜎 𝑦
𝐶. 𝐺
𝑇
𝑑
𝑏
2
3
𝑑 𝛼𝑑
2
3
𝛼𝑑
ൗ𝑑
2
𝐶 𝐶1
𝐶2
𝑇1
𝑇2
𝑇
𝐶
𝜎 𝑦
𝜎 𝑦 𝜎 𝑦
4
𝑑 1 + 𝛼
2
𝑃. 𝑁. 𝐴
ൗ𝑑
2
ൗ𝑑
2
𝑑 (1 − 𝛼)
2
𝑏 ×
𝑑 (1 + 𝛼)
2
+
1
2
𝜎 𝑦
𝛼 𝑑
2
𝑏 ×
2
3
𝛼 𝑑
𝑏𝑑2
4
1 − 𝛼2 +
2
3
𝛼2
𝑏𝑑2
4
1 −
𝛼2
3
= 𝜎 𝑦
𝑏𝑑2
6
3 − 𝛼2
2
using stress block:𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪

𝐸. 𝑁. 𝐴
𝜎 𝑦
𝐶. 𝐺
𝑇
𝑑
𝑏
2
3
𝑑 𝛼𝑑
2
3
𝛼𝑑
ൗ𝑑
2
𝐶 𝐶1
𝐶2
𝑇1
𝑇2
𝑇
𝐶
𝜎 𝑦
𝜎 𝑦 𝜎 𝑦
4
𝑑 1 + 𝛼
2
𝑃. 𝑁. 𝐴
ൗ𝑑
2
ൗ𝑑
2
𝑷𝑳𝑨𝑺𝑻𝑰𝑪
𝐶 = 𝑇 = 𝜎 𝑦
𝑑
2
𝑏 , 𝐴𝑟𝑚 =
𝑑
2
𝑀 𝑝 = 𝜎 𝑦
𝑑
2
𝑏 ×
𝑑
2
= 𝜎 𝑦
𝑏 𝑑2
4
= 𝜎 𝑦 𝑍 𝑝
using stress block:
𝑊ℎ𝑒𝑟𝑒: 𝑍 𝑝 =
𝑏 𝑑2
4

𝑓 =
𝑀 𝑝
𝑀 𝑦
=
𝜎 𝑦 𝑍 𝑝
𝜎 𝑦 𝑍 𝑒
=
𝑍 𝑝
𝑍 𝑒
Shape Factor (𝒇):
• It is the ratio of elastic to plastic moment capacity:
For rectangular section:
𝑓 =
𝑏 𝑑2
4
𝑏 𝑑2
6
=
6
4
= 1.5 𝑓 = 1.5
𝑓 = 1.698
𝑓 = 2.0
𝑓 ≈ 1.12 𝑡𝑜 1.15
𝑓 ≈ 1. 8
𝑓 ≈ 1. 0
Shapefactor(f)increasesasthematerial
inthemiddleofsectionincreases
𝑓 ≈ 1.27

Example (1):
• Find the plastic moment and shape factor for the shown section.
200 𝑚𝑚
200 𝑚𝑚
360𝑚𝑚20𝑚𝑚20𝑚𝑚
10 𝑚𝑚
4.2. Doubly Symmetric Sections

Example (1):
1- Elastic
𝐸. 𝑁. 𝐴
200 𝑚𝑚
200 𝑚𝑚
10 𝑚𝑚
200𝑚𝑚200𝑚𝑚
𝐼 =
10 × 3603
12
+2 ×
200 × 203
12
+ 200 × 20 × 1902
𝐼 = 327946666.7 𝑚𝑚4
𝑍 𝑒 =
327946666.7
200
= 1639733.333 𝑚𝑚3
𝑀 𝑦 = 𝜎 𝑦 𝑍 𝑒 = 1639733.333 𝜎 𝑦 𝑁. 𝑚𝑚

Example (1):
2- Plastic
𝐸. 𝑁. 𝐴
𝐶2
200 𝑚𝑚
𝜎 𝑦
𝑃. 𝑁. 𝐴
200 𝑚𝑚
10 𝑚𝑚
𝜎 𝑦
180𝑚𝑚
380𝑚𝑚
𝐶1
𝑇1
𝑇2

Example (1):
2- Plastic
𝐸. 𝑁. 𝐴
𝐶2
200 𝑚𝑚
𝜎𝑦
𝑃. 𝑁. 𝐴
200 𝑚𝑚
10 𝑚𝑚
𝜎𝑦
180𝑚𝑚
380𝑚𝑚
𝐶1
𝑇1
𝑇2
𝐶1 = 𝑇1 = 200 × 20 × 𝜎 𝑦 = 4000𝜎 𝑦 N
𝑍 𝑝 = 1844000 𝑚𝑚3
𝑓 =
𝑀 𝑝
𝑀𝑒
=
𝑍 𝑝
𝑍 𝑒
=
1844000
1639733.333
𝑓 = 1.124573101
𝐶2 = 𝑇2 = 10 × 180 × 𝜎 𝑦 = 1800𝜎 𝑦 N
𝑀 𝑝 = 𝐶1 𝐴𝑟𝑚1 + 𝐶2 𝐴𝑟𝑚2 = 4000 𝜎 𝑦 × 380 + 1800 𝜎 𝑦 × 180
𝑀 𝑝 = 1844000 𝜎 𝑦 𝑁. 𝑚
3- Shape Factor (f)

Example (2):
• Find the plastic moment and shape factor for the shown section.
160 𝑚𝑚200𝑚𝑚
10 𝑚𝑚
4.3. Sections Symmetric about Vertical Axis

Example (2):
1- Elastic
𝐴 = 160 × 16 + 10 × 184 = 4400 𝑚𝑚2
𝐴. 𝑦𝑒 = 160 × 16 × 8 + 10 × 184 × 108
𝑦𝑒 =
219200
4400
= 49.818 𝑚𝑚
𝐼𝑧 =
(160 − 10) × 163
3
+
10 × 2003
3
− 4400 × 49.8182
= 15951321.21 𝑚𝑚4
𝑀 𝑦 = 𝜎 𝑦 𝑍 𝑒 = 106213.3979 𝜎 𝑦 𝑁. 𝑚𝑚
160 𝑚𝑚
200𝑚𝑚
10 𝑚𝑚
49.818𝑚𝑚
𝐸. 𝑁. 𝐴
𝑦
𝑧
𝑍 𝑒 =
15951321.21
200 − 49.818
= 106213.3979 𝑚𝑚3

160 𝑚𝑚200𝑚𝑚
10 𝑚𝑚
𝐸. 𝑁. 𝐴
𝜎 𝑦
𝐶
𝑇
𝜎 𝑦
𝑦
𝑧
Example (2):
2- Plastic
To find plastic center, assume plastic neutral axis
(P.N.A) at bottom of flange:
𝐶 = 160 × 16 × 𝜎 𝑦 = 2560 𝜎 𝑦 , 𝑇 = 184 × 10 × 𝜎 𝑦 = 1840 𝜎 𝑦 𝑁
Since 𝐶 > 𝑇, the plastic neutral axis must be shifted upward inside
the flange.

160 𝑚𝑚200𝑚𝑚
10 𝑚𝑚
13.75𝑚𝑚
49.818𝑚𝑚
𝐸. 𝑁. 𝐴
𝜎 𝑦
𝐶1
𝑇2
𝜎 𝑦
𝑦
𝑧
𝑃. 𝑁. 𝐴
𝑇1
Example (2):
2- Plastic
The compression areas equal to tension area: Ac = At
160 × 𝑦𝑝 = 160 × 16 − 𝑦𝑝 + 10 × 184 → 𝑦𝑝 = 13.75 𝑚𝑚

Example (2):
2- Plastic
160 𝑚𝑚
200𝑚𝑚
10 𝑚𝑚
13.75𝑚𝑚
49.818𝑚𝑚
𝐸. 𝑁. 𝐴
𝜎𝑦
𝐶1
𝑇2
𝜎𝑦
𝑦
𝑧
𝑃. 𝑁. 𝐴
𝑇1
𝐶1 = 160 × 13.75 × 𝜎 𝑦 = 2200𝜎 𝑦 N
𝑍 𝑝 = 188950 𝑚𝑚3
𝑓 =
𝑀 𝑝
𝑀𝑒
=
𝑍 𝑝
𝑍 𝑒
=
188950
106213.3979
= 1.779
𝑇1 = 160 × 2.25 × 𝜎 𝑦 = 360𝜎 𝑦 N
𝑀 𝑝 = 2200 𝜎 𝑦 × 6.875 + 360 𝜎 𝑦 × 1.125 + 1840 𝜎 𝑦 × 94.25
= 188950 𝜎 𝑦 𝑁. 𝑚
3- Shape Factor (f)
𝑇2 = 10 × 184 × 𝜎 𝑦 = 1840𝜎 𝑦 N

𝜀 𝑦
𝐶. 𝐺
𝑑
𝑏
𝛼𝑑
𝜀 𝑦
𝜀 𝑦
𝜀 𝑦
ൗ𝑑
2
𝐸. 𝑁. 𝐴
𝜎 𝑦
𝐶. 𝐺
𝑇
𝑑
𝑏
2
3
𝑑 𝛼𝑑
2
3
𝛼𝑑
ൗ𝑑
2
𝐶 𝐶1
𝐶2
𝑇1
𝑇2
𝑇
𝐶
𝜎 𝑦
𝜎 𝑦 𝜎 𝑦
4
𝑑 1 + 𝛼
2
𝑃. 𝑁. 𝐴
ൗ𝑑
2
ൗ𝑑
2
ൗ𝑑
2 𝑦
𝜀
𝜅
𝜀
𝜀
𝑦
𝜀
𝜅𝑦
𝜀
Strain
Stress
𝜅 𝑦
5.1. Rectangular Sections
5. Moment-Rotation Relationship

𝜀 𝑦
𝐶. 𝐺
𝑑
𝑏
𝛼𝑑
𝜀 𝑦
𝜀 𝑦
𝜀 𝑦
ൗ𝑑
2
ൗ𝑑
2
𝜅 𝑦
𝑦
𝜀
𝜅
𝜀
𝜀
𝑦
𝜀
𝜅𝑦
𝜀
Strain
𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒, 𝜅 =
𝜀
𝑦
𝑬𝑳𝑨𝑺𝑻𝑰𝑪:
𝜅 𝑦 =
𝜀 𝑦
𝑑/2
= 2
𝜀 𝑦
𝑑
When the section is subjected to
yield moment 𝑀 𝑦
𝜅 =
1
𝑅
=
𝑀
𝐸𝐼
𝜅 𝑦 =
𝑀 𝑦
𝐸𝐼
𝑀
𝑀 𝑦
=
𝜅
𝜅 𝑦

𝜀 𝑦
𝐶. 𝐺
𝑑
𝑏
𝛼𝑑
𝜀 𝑦
𝜀 𝑦
𝜀 𝑦
ൗ𝑑
2
ൗ𝑑
2
𝜅 𝑦
𝑦
𝜀
𝜅
𝜀
𝜀
𝑦
𝜀
𝜅𝑦
𝜀
Strain
𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪:
𝜅 =
𝜀 𝑦
𝛼 𝑑/2
= 2
𝜀 𝑦
𝛼 𝑑
For moments applied beyond the yield moment, the curvature can be
found by noting that the yield strain, 𝜀 𝑦 , occurs at a distance from
the neutral axis of α 𝑑/2, giving:
Thus, the ratio curvature to yield curvature is:
𝜅
𝜅 𝑦
= ൘
2 𝜀 𝑦
𝛼 𝑑
2 𝜀 𝑦
𝑑
=
1
𝛼
→ 𝛼 =
𝜅 𝑦
𝜅

𝑀
𝑀 𝑦
=
𝜎 𝑦
𝑏𝑑2
6
3 − 𝛼2
2
𝜎 𝑦
𝑏𝑑2
6
=
3 − 𝛼2
2
Also, the ratio of elasto-plastic moment to yield moment is:
If we now substitute the value 𝛼 =
𝜅 𝑦
𝜅
we find:
𝑀
𝑀 𝑦
=
1
2
3 −
𝜅 𝑦
𝜅
2
= 1.5 − 0.5
𝜅
𝜅 𝑦
−2

0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
0 1 2 3 4 5 6 7 8
/
/
𝑀
𝑀 𝑦
= 1.5 − 0.5
𝜅
𝜅 𝑦
−2
𝑀
𝑀 𝑦
=
𝜅
𝜅 𝑦
𝑑
𝑏
𝑦
𝑧
𝑀 = 𝑀 𝑦
𝑀 = 𝑀 𝑝
𝑓 =

There are some important observations to be made from
this graph:
• To reach the plastic moment capacity of the section
requires large curvatures. Thus the section must be
ductile.
• The full cross-section plasticity associated with the
plastic moment capacity of a section can only be reached
at infinite curvature (or infinite strain). Since this is
impossible, we realize that the full plastic moment
capacity is unobtainable.

To show that the idea of the plastic moment capacity of
section is still useful. Firstly we note that strain hardening
in mild steel begins to occur at a strain of about (10 𝜀 𝑦).
At this strain, the corresponding moment ratio is:
𝑀
𝑀 𝑦
= 1.5 − 0.5 10 −2 = 1.495
Since this is about 99.7% of the plastic moment capacity,
we see that the plastic moment capacity of a section is a
good approximation of the section’s capacity.
These calculations are based on a ductility ratio of 10. This
is about the level of ductility a section requires to be of
use in any plastic collapse analysis.

Lastly, for other cross-section shapes we have the
moment-curvature relations shown in the following figure.
𝒇𝒓𝒐𝒎 [𝑪𝒐𝒍𝒊𝒏 𝑪𝒂𝒑𝒓𝒂𝒏𝒊]

𝐸. 𝑁. 𝐴𝐶. 𝐺
𝑑
𝑏
ൗ𝑑
2
𝑇
𝐶
𝜎 𝑦
𝜎 𝑦
𝑃. 𝑁. 𝐴
ൗ𝑑
2
ൗ𝑑
2
𝐶 = 𝑇 = 𝜎 𝑦
𝑑
2
𝑏
+
2 𝜎 𝑦
𝐶` =
𝜎 𝑦
𝜎 𝑦
𝛽 𝑑
𝐶` = 2 𝜎 𝑦 𝛽 𝑑 𝑏 , 𝐴𝑟𝑚 =
𝛽 𝑑
2
the plastic ‘squash load’ of the section is given by: 𝑃𝑐 = 𝜎 𝑦 𝑏 𝑑
𝑇ℎ𝑢𝑠, 𝑃 = 2 𝛽𝑃𝑐
𝑃 = 𝐶 − 𝑇 + 𝐶` = 𝐶`
𝑀 𝑝𝑐 = 𝑀 𝑝 − 𝑃
𝛽 𝑑
2
The plastic moment is given by:
6. Effect of Axial Force on the Plastic Moment Capacity

𝑈𝑠𝑖𝑛𝑔, 𝑀 𝑝 = 𝜎 𝑦
𝑏 𝑑2
4
𝑀 𝑝𝑐 = 𝜎 𝑦
𝑏 𝑑2
4
− 2 𝜎 𝑦 𝛽 𝑑 𝑏
𝛽 𝑑
2
= 𝜎 𝑦
𝑏 𝑑2
4
1 − 4𝛽2
𝑀 𝑝𝑐 = 𝑀 𝑝 1 − 4𝛽2
𝛽 𝑑
2
The plastic moment is given by:
𝑁𝑜𝑡𝑖𝑛𝑔 𝑡ℎ𝑎𝑡, 2 𝛽 = 𝑃/𝑃𝑐
𝑀 𝑝𝑐
𝑀 𝑝
= 1 − 4𝛽2 = 1 −
𝑃
𝑃𝑐
2
The interaction equation is
given by:
𝑀 𝑝𝑐
𝑀 𝑝
+
𝑃
𝑃𝑐
2
= 1 0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
P/Pc
Mpc/Mp
𝑀 𝑝𝑐
𝑀 𝑝
+
𝑃
𝑃𝑐
2
= 1
𝑑
𝑏
𝑦
𝑧

Example (3):
• Deduce and plot the interaction equation between axial and
bending of the shown section.
200 𝑚𝑚
200 𝑚𝑚
10 𝑚𝑚

Example (3):
Crushing Load: 𝑃𝑐 = 𝐴. 𝜎 𝑦 = 10 × 360 + 2 × 200 × 20 𝜎 𝑦 = 11600 𝜎 𝑦 𝑁
𝐸. 𝑁. 𝐴
𝐶2
200 𝑚𝑚
𝜎 𝑦
𝑃. 𝑁. 𝐴
200 𝑚𝑚
10 𝑚𝑚
𝜎 𝑦
𝐶1
𝑇1
𝑇2
𝑃𝑐
Pure Moment Pure Axial Load
𝜎 𝑦
𝜎 𝑦
Plastic Moment (deduced before): 𝑀 𝑝 = 1844000 𝜎 𝑦 𝑁. 𝑚

Example (3):
Stress distribution: 1) 𝛽 𝑑 ≤ 180 𝑚𝑚
𝐸. 𝑁. 𝐴
𝐶2
200 𝑚𝑚
𝜎 𝑦
𝑃. 𝑁. 𝐴
200 𝑚𝑚
10 𝑚𝑚
𝜎 𝑦
200𝑚𝑚200𝑚𝑚 𝐶1
𝑇1
𝑇2
𝛽 𝑑
2 𝜎 𝑦
Pure Moment Effect of Axial
Force (P)
+ =
𝜎 𝑦
𝜎 𝑦
𝐶`
Combined
Effect

Example (3):
Modified Plastic Moment:
𝑀 𝑝𝑐
𝑀 𝑝
=
1844000 − 1600000 𝛽2
1844000
= 1 −
400
461
𝛽2
𝑀 𝑝𝑐
𝑀 𝑝
= 1 −
400
461
29
20
𝑃
𝑃𝑐
2
= 1 −
841
461
𝑃
𝑃𝑐
2
𝑏𝑢𝑡 𝛽 =
29
20
𝑃
𝑃𝑐
𝑀 𝑝𝑐
𝑀 𝑝
+
841
461
𝑃
𝑃𝑐
2
= 1Interaction Equation:
Load: 𝑃 = 2𝜎 𝑦 𝛽 𝑑 𝑡 𝑤 = 2 × 400 × 10 = 8000 𝛽 𝜎 𝑦 𝑁
Load ratio:
𝑃
𝑃𝑐
=
8000 𝛽 𝜎 𝑦
11600 𝜎 𝑦
=
20
29
𝛽 → 𝛽 =
29
20
𝑃
𝑃𝑐
𝛽𝑑
2
𝑀 𝑝𝑐 = 𝑀 𝑝 −
400
2
𝛽 × 8000 𝛽 𝜎 𝑦 = 1844000 − 1600000 𝛽2
𝜎 𝑦

Example (3):
Stress distribution: 2) 𝛽 𝑑 ≥ 180 𝑚𝑚
𝐸. 𝑁. 𝐴
200 𝑚𝑚
𝜎 𝑦
𝑃. 𝑁. 𝐴
200 𝑚𝑚
10 𝑚𝑚
𝜎 𝑦
𝛽 𝑑
2 𝜎 𝑦
Pure Axial
Force (Pc)
Effect of Axial
Force (P)
𝜎 𝑦
𝜎 𝑦
𝐶`
Combined
Effect
𝑃𝑐
=+

Example (3):
Modified Plastic Moment:
Load:
Load ratio:
𝑃
𝑃𝑐
= 1 −
80000 1 − 2 𝛽 𝜎 𝑦
11600 𝜎 𝑦
= 1 −
200
29
1 − 2 𝛽
𝑃
𝑃𝑐
=
400
29
𝛽 −
171
29
→
𝑀 𝑝𝑐 = 𝐶` × 𝐴𝑟𝑚 = 2𝜎 𝑦 𝑏
𝑑
2
− 𝛽 𝑑 ×
1
2
𝑑
2
+ 𝛽 𝑑
𝑃 = 𝑃𝑐 − 2𝜎 𝑦
𝑑
2
− 𝛽 𝑑 𝑏 = 𝑃𝑐 − 𝜎 𝑦 × 400 × 200 1 − 2 𝛽
𝑃 = 𝑃𝑐 − 80000 1 − 2 𝛽 𝜎 𝑦 𝑁
𝛽 =
29
400
𝑃
𝑃𝑐
+
171
400
𝑀 𝑝𝑐 = 𝐶` × 𝐴𝑟𝑚 = 𝜎 𝑦
𝑏 𝑑2
4
1 − 4 𝛽2
= 8 × 106
1 − 4 𝛽2
𝜎 𝑦
𝑀 𝑝𝑐
𝑀 𝑝
=
8 × 106
1 − 4 𝛽2
𝜎 𝑦
1844000𝜎 𝑦
=
2000
461
1 − 4 𝛽2

Example (3):
This yields:
But 𝛽 =
29
400
𝑃
𝑃𝑐
+
171
400
𝑀 𝑝𝑐
𝑀 𝑝
=
2000
461
1 − 4
29
400
𝑃
𝑃𝑐
+
171
400
2
𝑀 𝑝𝑐
𝑀 𝑝
=
10759
9220
−
9918
9220
𝑃
𝑃𝑐
−
841
9220
𝑃
𝑃𝑐
2
9220
10759
𝑀 𝑝𝑐
𝑀 𝑝
+
342
371
𝑃
𝑃𝑐
+
29
371
𝑃
𝑃𝑐
2
= 1
The interaction equation:

Example (3):
There are 2 interaction equations:
𝑀 𝑝𝑐
𝑀 𝑝
= 1 −
841
461
𝑃
𝑃𝑐
2
= 1 −
841
461
9
29
2
=
380
461
= 0.8243
𝑃
𝑃𝑐
=
20
29
𝛽 =
20
29
×
180
400
=
9
29
= 0.3103
• The first one is valid for 𝛽𝑑 = 0 𝑡𝑜 180
• The other is valid for 𝛽𝑑 = 180 𝑡𝑜 200
Both equations are valid for 𝛽𝑑 = 180 , 𝑖. 𝑒. 𝛽 = Τ180 400
By substituting into
By substituting into first interaction equation
Same result can be obtained by substituting into second equation:
𝑀 𝑝𝑐
𝑀 𝑝
=
10759
9220
−
9918
9220
×
9
29
−
841
9220
×
9
29
2
=
380
461
= 0.8243

0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
P/Pc
Mpc/Mp
Example (3):
𝟗𝟐𝟐𝟎
𝟏𝟎𝟕𝟓𝟗
𝑴 𝒑𝒄
𝑴 𝒑
+
𝟑𝟒𝟐
𝟑𝟕𝟏
𝑷
𝑷 𝒄
+
𝟐𝟗
𝟑𝟕𝟏
𝑷
𝑷 𝒄
𝟐
= 𝟏
𝑴 𝒑𝒄
𝑴 𝒑
+
𝟖𝟒𝟏
𝟒𝟔𝟏
𝑷
𝑷 𝒄
𝟐
= 𝟏
𝟎. 𝟑𝟏𝟎𝟑
𝟎.𝟖𝟐𝟒𝟑
200 𝑚𝑚
200 𝑚𝑚
10 𝑚𝑚

Example (3):
𝟗𝟐𝟐𝟎
𝟏𝟎𝟕𝟓𝟗
𝑴 𝒑𝒄
𝑴 𝒑
+
𝟑𝟒𝟐
𝟑𝟕𝟏
𝑷
𝑷 𝒄
+
𝟐𝟗
𝟑𝟕𝟏
𝑷
𝑷 𝒄
𝟐
= 𝟏
𝑴 𝒑𝒄
𝑴 𝒑
+
𝟖𝟒𝟏
𝟒𝟔𝟏
𝑷
𝑷 𝒄
𝟐
= 𝟏
𝑷
𝑷 𝒄
> 𝟎. 𝟏𝟓:
𝑴 𝒑𝒄
𝑴 𝒑
= 𝟏. 𝟏𝟖 𝟏 −
𝑷
𝑷 𝒄
𝑷
𝑷 𝒄
≤ 𝟎. 𝟏𝟓:
𝑴 𝒑𝒄
𝑴 𝒑
= 𝟏. 𝟎
𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆:
𝒇𝒓𝒐𝒎 [𝑪𝒐𝒍𝒊𝒏 𝑪𝒂𝒑𝒓𝒂𝒏𝒊]
200 𝑚𝑚
200 𝑚𝑚
10 𝑚𝑚

𝑧
𝑃
𝑀𝑒𝑝
𝛼 𝑑
𝑀 𝑝
𝑀 𝑦
𝑀 = 0
𝑙 𝑝
𝐿/2
𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑒𝑙𝑎𝑠𝑡𝑜 − 𝑝𝑙𝑎𝑠𝑡𝑖𝑐
𝑝𝑙𝑎𝑠𝑡𝑖𝑐
𝑑
𝑃 < 𝑃𝑦
𝑃 = 𝑃𝑦
𝑃 = 𝑃𝑝
𝐵. 𝑀. 𝐷
𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑧𝑜𝑛𝑒
𝐿/2
7.1. Simply-Supported Beam Subjected to Midspan Point Load
7. Plastic Hinge Formation

The bending moment at center of beam is given by:
𝑀𝑐 =
𝑃𝐿
4
→ 𝑃 =
4𝑀𝑐
𝐿
The bending moment at first yield (at center of beam) is given by:
𝑀𝑐 = 𝑀 𝑦 =
𝑃𝑦 𝐿
4
→ 𝑃𝑦 =
4𝑀 𝑦
𝐿
The collapse load occurs when the moment at the center reaches the
plastic moment capacity:
𝑀𝑐 = 𝑀 𝑝 =
𝑃𝑝 𝐿
4
→ 𝑃𝑝 =
4𝑀 𝑝
𝐿

The ratio collapse to yield load is:
𝑃𝑃
𝑃𝑦
=
4𝑀 𝑝
𝐿
4𝑀 𝑦
𝐿
=
𝑀 𝑝
𝑀 𝑦
𝑏𝑢𝑡,
𝑀 𝑝
𝑀 𝑦
= 𝑓 𝑡ℎ𝑢𝑠,
𝑃𝑝
𝑃𝑦
= 𝑓
This is a general result:
the ratio of collapse load to first yield load is the shape factor of the
member, for statically determinate prismatic structures.

Shape of the Plastic Hinge:
We are also interested in the plastic hinge, and the zone of Elasto-
plastic bending. As can be seen from the diagram, the plastic material
zones extend from the center out to the point where the moment
equals the yield moment. Using similar triangles from the bending
moment diagram at collapse, we see that:
𝑀 𝑃
𝐿
=
𝑀 𝑝 − 𝑀 𝑦
𝑙 𝑝
=
𝑀 𝑝 − 𝑀𝑒𝑝
2𝑧
Equating the first two equations gives:
𝑙 𝑝 = 𝐿
𝑀 𝑃
= 𝐿 1 −
𝑀 𝑦
𝑀 𝑝
= 𝐿 1 −
1
𝑓
For a beam with a rectangular cross section (𝑓 = 1.5) the plastic hinge
extends for a length:
𝑙 𝑝 = 𝐿 1 −
1
1.5
=
𝐿
3

Lastly, the shape of the hinge follows from the first and third equation:
𝑀 𝑃
𝐿
=
2𝑧
→
𝑧
𝐿
=
1
2
1 −
𝑀𝑒𝑝
𝑀 𝑝
𝑧
𝐿
=
1
2
1 −
𝜎 𝑦
𝑏𝑑2
6
3 − 𝛼2
2
𝜎 𝑦
𝑏𝑑2
4
=
1
2
1 −
1
3
3 − 𝛼2
𝑧
𝐿
=
𝛼2
6
, 𝑎𝑡 𝛼 = 1, 𝑧 =
𝑙 𝑝
2
=
𝐿
6
, 𝑙 𝑝 =
𝐿
3
From expressions for the elasto-plastic and plastic moments, we have:
𝐹𝑖𝑛𝑎𝑙𝑙𝑦, 𝛼 =
6𝑧
𝐿

𝑧
𝑤
𝑀𝑒𝑝
𝛼 𝑑
𝑀 𝑝
𝑀 𝑦
𝑀 = 0
𝑙 𝑝
𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑒𝑙𝑎𝑠𝑡𝑜 − 𝑝𝑙𝑎𝑠𝑡𝑖𝑐
𝑝𝑙𝑎𝑠𝑡𝑖𝑐
𝑑
𝑤 < 𝑤 𝑦
𝑤 = 𝑤 𝑦
𝑤 = 𝑤 𝑝
𝐵. 𝑀. 𝐷
𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑧𝑜𝑛𝑒
𝐿/2 𝐿/2
7.2. Simply-Supported Beam Subjected to Uniform Load

The bending moment at center of beam is given by:
𝑀𝑐 =
𝑤𝐿2
8
→ 𝑤 =
8𝑀𝑐
𝐿2
The bending moment at first yield (at center of beam) is given by:
𝑀𝑐 = 𝑀 𝑦 =
𝑤 𝑦 𝐿2
8
→ 𝑤 𝑦 =
8𝑀 𝑦
𝐿2
The collapse load occurs when the moment at the center reaches the
plastic moment capacity:
𝑀𝑐 = 𝑀 𝑝 =
𝑤 𝑝 𝐿2
8
→ 𝑤 𝑝 =
8𝑀 𝑝
𝐿2

The ratio collapse to yield load is:
𝑤 𝑃
𝑤 𝑦
=
8𝑀 𝑝
𝐿2
8𝑀 𝑦
𝐿2
=
𝑀 𝑝
𝑀 𝑦
𝑏𝑢𝑡,
𝑀 𝑝
𝑀 𝑦
= 𝑓 𝑡ℎ𝑢𝑠,
𝑤 𝑝
𝑤 𝑦
= 𝑓
This is a general result:
the ratio of collapse load to first yield load is the shape factor of the
member, for statically determinate prismatic structures.

We are also interested in the plastic hinge, and the zone of Elasto-
plastic bending. As can be seen from the diagram, the plastic material
zones extend from the center out to the point where the moment
equals the yield moment. Using parabolic proportion from the bending
moment diagram at collapse, we see that:
𝑀 𝑃
𝐿
2
2 =
𝑙 𝑝
2
2 =
𝑧2
Equating the first two equations gives:
𝑙 𝑝
2 = 𝐿2
𝑀 𝑃
= 𝐿2 1 −
𝑀 𝑦
𝑀 𝑝
= 𝐿2 1 −
1
𝑓
→
For a beam with a rectangular cross section (𝑓 = 1.5) the plastic hinge
extends for a length:
𝑙 𝑝 = 𝐿 1 −
1
𝑓
𝑙 𝑝 = 𝐿 1 −
1
1.5
=
𝐿
3
𝑙
𝑥
𝑏
𝑎
𝑎
𝑙2
=
𝑏
𝑥2

Lastly, the shape of the hinge follows from the first and third equation:
𝑧
𝐿
=
1
2
1 −
𝜎 𝑦
𝑏𝑑2
6
3 − 𝛼2
2
𝜎 𝑦
𝑏𝑑2
4
=
1
2
1 −
1
3
3 − 𝛼2
𝑧
𝐿
=
𝛼
2 3
, 𝑎𝑡 𝛼 = 1, 𝑧 =
𝑙 𝑝
2
=
𝐿
2 3
, 𝑙 𝑝 =
𝐿
3
From expressions for the elasto-plastic and plastic moments, we have:
𝐹𝑖𝑛𝑎𝑙𝑙𝑦, 𝛼 = 2 3
𝑧
𝐿
𝑀 𝑃
𝐿
2
2 =
𝑧2
→
𝑧
𝐿
=
1
2
1 −
𝑀𝑒𝑝
𝑀 𝑝
Which is an equation of straight line

• Colin Caprani “Plastic Analysis 3rd Year Structural
Engineering”, 2009, Monash University (Australia).
• T.H.G. Megson “Structural and Stress Analysis”, 2nd ed.,
2005, Elsevier.
• “Chapter 35: PLASTIC ANALYSIS”, Version 2, INSTITUTE
FOR STEEL DEVELOPMENT & GROWTH,
http://www.steel-insdag.org.
REFERENCES

Part (1) basics of plastic analysis

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