CHAPTER 3

BENDING AND TORSION UNDER

PLASTIC CONDITIONS
MOMENT UNDER PLASTIC CONDITION

1
 INTRODUCTION
Until now we have learnt that when designing engineering structures, the allowable stresses
should not exceed the elastic limit of the material (in tension or compression). This is known as
designing based on elastic theory or elastic design.

The allowable stress (σallowable) is taken to be the yield stress of the material divided by a
convenient factor of safety (usually based on design codes or experience).

Beyond the elastic limit, plastic deformation occurs.

2
 INTRODUCTION
Failure of actual structures is a complex phenomenon because plastic yielding/deformation begins
at the extreme fibres and proceeds towards the neutral axis.

This calls for the extension of elastic theory to account for the plastic deformation of structural
materials.
Elastic vs. Plastic Analysis
Consider a specimen of structural steel under a tensile test. The relationship between stress and
strain is shown below.
ultimate σ
σ upper yield stress upper yield
point
lower yield
strain hardening

plastic zone σy
lower yield point

ε 3
ε
 Elastic vs. Plastic Analysis
σ
upper yield
lower yield

elastic plastic σy
zone zone

In elastic analysis (e.g., slope-deflection of beams, failure of columns, etc.), the focus is on the
elastic region. That is, design is based on the assumption that there is no yield anywhere in the
structure.
In plastic analysis, the focus is on the both the elastic and plastic regions. That is, the design
must account for plastic yielding or deformation of the structure.
Advantages of Plastic Design:
(a) More suitable for steel structures that have more deformations
(b) Results in more economical structures 4
 Plastic Theory of Bending
Consider a simply supported rectangular beam subjected to a point load, W (see figure a).
W W yielded/plastic
zones

(a) (b)

As the applied load is gradually increased, starting from zero, the beam is stressed first in a
purely elastic manner until the elastic limit is reached.
At the elastic limit, the stresses in the upper and lower edges of the central section will have
reached the yield stress, σy.
Further increase of the load will cause the yield zones to spread and move towards the neutral
axis until the beam rotates freely about the load in a manner similar to a hinge. At this stage, a
plastic hinge is said to have developed. 5
 Assumptions in the Plastic Theory
 The material exhibits marked yield, and can undergo considerable strain during
plastic deformation without further increase in stress.

 The yield stress is taken to be the lower yield and is the same in tension and
compression

 Transverse cross-sections remain plane, so that strain is proportional to the distance

from the neutral axis, though in the plastic region stress will be constant, and not
proportional to strain.

 When a plastic hinge has developed at any cross-section the moment of resistance
at that point remains constant until collapse of the whole structure takes place due
to the formation of the required number of further plastic hinges at other points.

6
 Moment of Resistance at the Plastic Hinges
Our objective is to determine the moment of resistance or the moment required to produce the
plastic hinge.

7
 Moment of Resistance at the Plastic Hinges
Rectangular Section:
The total moment of resistance ( Fig b)
M  M elastic  M plastic  16  y bd  2h    y bhd  h 
2

In an elastic range, h = 0, (Fig a)

I   bd 2 
My  y   y
 y  6 

   
b
For fully plastic, h = d/2
h

3 y bd 2
 1
Mp    4  y bd 2  1.5M y
2  6  d
The ratio Mp\My is called the Mp
shape factor , S s  1 .5 h
My 8
 Example 1
A steel bar of rectangular section 72 mm by 30 mm is used as a simply supported beam on a
span of l.2 m and loaded at mid-span. If the yield stress is 280 N/mm2 and the long edges of the
section are vertical, find the load when yielding first occurs.
Assuming that a further increase in load causes yielding to spread inwards towards the neutral
axis, with the stress in the yielded part remaining at 280 N/mm2, find the load required to cause
yielding for a depth of 12 mm at the top and bottom of the section at mid-span, and find the
length of beam over which yielding has occurred.

Solution
If Wy is the load at first yield, then:

 bd 2 
M y    y 
30 72  280 
2
 7257600 Nmm...(i)
 6  6
M y  300Wy ...(ii)
From (i) and (ii) Wy  24.2 kN 9
 Example 1 (continued)
The outer 12 mm on each side of the neutral axis being under constant stress of 280 N/mm2
M  6  y bd  2h    y bhd  h 
with no drop of stress at yield. 1 2

The moment of resistance

 16 280 30 48  280 30 12 60 
2

M  300W  9273600 W  31kN

 3225600  6048000
First yield
W y 0.3  0.3(24.2)  7.26kN  9273600

Distance x from either end under the central load W

1
2 Wx  7.26  x  0.468m
The length of the beam: 1.2  2x  0.264m 10
 Moment of Resistance at the Plastic Hinges
I-Section b

In an elastic range, (Fig a)  bd 3 b1d13  2  1

b1
M y   y     2

 12 12  d  d
d1

For fully plastic, (Fig c)

 bd 2 b1 d12 
M p   y   
 4 4 
Mp  bd 2  b1 d12 
The ratio Mp\My is called the shape factor , S S  1.5d  
My  bd 3
 b d
1 1
3

The shape factor will vary slightly with the proportions of flange to web, an average value
11
being about 1.15, as illustrated by the example below.
 Example 2
A 300 mm by 125 mm I-beam has flanges 13 mm thick and web 8.5 mm thick. Calculate the
shape factor and the moment of resistance in the fully plastic state. Take σy =250 N/mm2 and Ix
=85 x 106 mm4

Solution
At fist yield,  
M y  I y  y  85.106 150 250  141x106 Nmm  141kNm

  250
 
2 2
b d
In the fully plastic,
M p   y 
bd
 1 1 
 125300 2
 116.5 274 2
 156kNm
 4 4  4

M p 156
The shape factor S   1.11
M y 141 12
 Moment of Resistance at the Plastic Hinges
General Cross Section
If A is the total area of cross-section, then it is clear that for pure bending in the fully plastic
state the "neutral axis" must divide the area into equal halves.

If the centroids of these halves are G1 and G2 at a distance y1 +y2 apart, then

M p  12  y A  y1  y2 
But at first yield M y  Z y
where Z is the section modulus.
Mp A y1  y 2 
Hence
S 
My 2Z 13
 Example 3
Find the shape factor for a 150 mm by 75 mm channel in pure bending with the plane of
bending perpendicular to the web of the channel. Take, Z =12,000 mm3.

Solution
150
150

6.25
A1 6.25
G1

y1
P P

75 75
y2

A2 A3
G2 h
9.5
9.5

14
 Example 3 (continued)
Area of the channel Let the centroids of these halves be G1
A2  A3  9.5(75  6.25)  653 mm 2 and G2 at a distance y1 +y2 apart.
A1  150(6.25)  937.5 mm 2
Therefore,
A1  2 A2  9.5(75  6.25)  937.5  2653  2243 mm
29.5h  1121.5
2

Neutral axis that divides the area into equal

halves  h  59mm
1
2 A 1
2
2243  1121.5 mm 2
Since, A1  12 A y1  h 2  29.5mm
Then, the "neutral axis will pass through A2
and A3 labeled PP as shown in the Fig
15
 Example 3 (continued)
1506.2516  3.125  2 9.59.754.875
y 2   11.8mm
1506.25  29.59.75

The shape factor

A y1  y 2  224341.76 
S    2.23
2Z 221000 

16
TORSION UNDER PLASTIC CONDITION

17
 Assumptions in the Plastic Theory
A plane cross section of the shaft remains plane
when in the plastic state. Shear
Stress
Radius remains straight 

ry

r0

Shear Strain  18
 Torque of Resistance at the Plastic Hinges
The total torque of resistance  y ry3
T1 
2
2
r0

ry 3

T2   2r y .rdr   y r03  ry3  2 
3
r
T  T1  T2   y r0 1  3
3
y


3  4r0 
 
In an elastic range, r0 = ry,
 y ry 3

Ty 
2
For fully plastic, ry = 0
2
T p   y r03

3 19
 Torque of Resistance at the Plastic Hinges
When the fibres at the outer surface of the shaft are about to become plastic, the angle of twist
of the shaft is given by
 yL  y ry  yL
y   
Gr0  r0 Gry
L the length of the shaft
G Modulus of Rigidity
The shape factor , S
For elastoplastic condition s  p  y  4 3
 y 
3

T   y r03 1  
2 1 
4 

3    
  20
 Example 4
A mild steel shaft 40 mm in diameter and 250 mm in length is subjected to an overload torque
of 1800 Nm which caused shear yielding of the shaft, 120 MPa. Determine the radial depth to
which plasticity has penetrated and the angle of twist. Take G = 80 GPa

Solution   
3
  3

The Torque
2 3
r
1 y   2
 6

T   y r0 1  4   1800   120.10 0.02 1 
3 r y
3
3 
  r0   3  40.02 
From which ry  15mm

Hence depth of plastic deformation is 5mm

 
ry 

y

 
120 10 6
 3

The shear strain at ry==15mm is L G  

80 10 9
1 . 5 x10

L 0.00150.25
    0.025rad  1.43
21
0

ry 0.015
 Further Examples
1. (a) A rectangular section beam, 50 mm wide by 20 mm deep, is used as a simply supported
beam over a span of 2 m with the dimension vertical. Determine the value of the central
concentrated load which will produce initiation of yield at the outer fibres of the beam.
(b) If the central load is then increased by 10%, find the depth to which yielding will take place
at the centre of the beam span.
(c) Over what length of beam will yielding then have taken place?
(d) What are the maximum deflections for each load case?
For steel σy in simple tension and compression = 225 MPa and E = 206.8 GPa.

2. A solid circular shaft of diameter 50 mm and length 300 mm is subjected to a gradually

increasing torque T. The yield stress in shear for the shaft material is 120 MPa and, up to the
yield point, the modulus of rigidity is 80 GPa.
(a) Determine the value of T and the associated angle of twist when the shaft material first
yields
(b) If, after yielding, the stress is assumed to remain constant for any further increase in strain,
determine the value of T when the angle of twist is increased to twice that at yield.
22