Lecture 11

Plastic analysis of slabs

Yield-line and strip methods
Print version Lecture on Theory of Elasticity and Plasticity

Dr. D. Dinev, Department of Structural Mechanics, UACEG

11.1

Contents
1 Yield-line method 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Yield lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Ultimate moment of resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.4 Analysis by virtual work principle . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Minimum load principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Strip method 7
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Basic principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.3 Choise of load distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 11.2

1 Yield-line method
1.1 Introduction
Yield line theory of slabs

Introduction

1
 • The method for a limit analysis of RC slabs known as yield line theory was initiated by
Ingerslev (1921) and extended by Johansen (1932)
11.3

Yield line theory of slabs

Introduction
• The ultimate load is estimated by postulating a collapse mechanism that is compatible with
the BCs (upper-bound approach)
• The moment at the plastic hinge lines are the ultimate moment of resistance of the section
• The ultimate load is determined using the principle of virtual work and it is either correct
or too high (upper-bound theorem)

Note
• Thus all possible collapse mechanisms must be examined to ensure that the load-carrying
capacity is not overestimated
11.4

1.2 Yield lines

Yield line theory of slabs

Yield lines
• A slab is assumed to collapse at its ultimate load trough a system of nearly straight lines,
which are called yield lines
• These yielad lines divide the slab into a number of panels
• The pattern of yield lines and panels is termed a collapse mechanism
11.5

Yield line theory of slabs

Yield lines
• When a collapse mechanism has developed, the plastic deformations along the yield lines
are much greater than the elastic deformations of the panels between the yield lines
• The theory assumes that the panels are plane
11.6

2
Yield line theory of slabs

Yield lines – basic rules

• The yield lines must be straight lines forming axis of rotation of the segments
11.7

Yield line theory of slabs

Yield lines – basic rules
• The supports acts as axes of rotation
• An axis of rotation will pass over a column
11.8

Yield line theory of slabs

Yield lines – basic rules
• The yield line must pass through the intersection of the axes of rotation of the adjacent
panels
11.9

1.3 Ultimate moment of resistance

Yield line theory of slabs
Ultimate moment of resistance
• For a yield line that runs orthogonal to the reinforcement the ultimate moment of resistance
per unit width is
 
As fy
Mu = φ As fy d −
1.7 fc0
where As is area of the reinforcement per unit width
11.10

Yield line theory of slabs

Ultimate moment of resistance

3
 • In the usual case the reinforcement bars are orthogonal to each other and not coincide with
a general yield line, then we may apply the Johansen’s yield criterion
• ∑ mn = 0 gives

mn (a) = mx sin α(a sin α) + my cos α(a cos α)

mn = mx sin2 α + my cos2 α
11.11

Yield line theory of slabs

Ultimate moment of resistance
• ∑ mt = 0 gives

mt (a) = mx cos α(a sin α) + my sin α(a cos α)

mt = (my − mx )sinα cos α

• When mx = my = m, thus mt = 0 the slab is isotropically reinforced

• What is the slab with mx 6= my ?!?
11.12

1.4 Analysis by virtual work principle

Yield line theory of slabs
Virtual work principle
• Suppose that a rigid body is in equilibrium under the action of a system of forces
• If the body is given a small arbitrary displacements, consistent with the BCs, the sum of the
work done by the forces (force times its corresponding displacement) will be zero because
the resultant force is zero
• The principle states: If a body that is in static equilibrium under the system of forces
is given a virtual displacement, the sum of the virtual work by the forces is zero
11.13

Yield line theory of slabs

Analysis by virtual work principle
• The virtual work W due to an applied load qu is
Z
W (qu ) = qu δ (x, y)dA = ∑ Qu ∆
A

where Qu is load resultant on the panel and ∆ is the displacement of its centroid

Hint
• The displacement of the center of gravity ∆ of a triangle abc is

1
∆ = (δa + δb + δc )
3
11.14

4
Yield line theory of slabs

Ultimate moment of resistance

• The virtual work due to an ultimate moment moments
U(m) = ∑ mL cos αθ
where m is the ultimate moment normal to the yield line, θ is the rotation and L is the yield
line length
11.15

Yield line theory of slabs

Analysis by virtual work principle
• The equilibrium is U = W or ∑ mL cos αθ = ∑ Qu ∆
• Since most slabs are rectangular and the reinforcement is orthogonal we know the mx and
my and it is easier to deal with the directional components of the internal work
• For an arbitrary yield line we have
U = ∑ mL cos αθ = ∑ mx θx y0 + ∑ my θy x0
where θx and θy are components of θn ; x0 and y0 are x and y projections of L
• Therefore

∑ Qu ∆ = ∑ mx θx Ly + ∑ my θy Lx
11.16

1.5 Minimum load principle

Yield line theory of slabs

Minimum load principle

• In most cases a yield line pattern cannot be drown without unknown dimensions locating
the yield line position
• The unknown dimensions must be included in the virtual work equation
• The equation for ultimate load has the form of Qu = f (x1 , x2 , . . . , xn )
11.17

5
Yield line theory of slabs

Minimum load principle

• Since the upper-bound approach is used the values for xn required those values that give
the minimum value for Qu and may be found by solving a set of equations

∂ Qu ∂ Qu
= 0, ... =0
∂ x1 ∂ xn
• The values for x1 , x2 , . . . , xn are substituted back into the ultimate load equation to obtain
the minimum Qu
11.18

Yield line theory of slabs

Example 1
• Consider a rectangular slab with an orthotropic reinforcement
• For the given pattern find the optimum position of the yield lines α =? and the ultimate
load qu =?
11.19

Yield line theory of slabs

6
Example 1
• The yield line pattern
11.20

2 Strip method
2.1 Introduction
Introduction
Àpplication of yield-line method- cons.
• The yield-line analysis is an upper-bound approach and thus, iff in error, will be so on the
unsafe side
• To apply it is necessery to assume that the reinforcement is known over the slab
• Therefore the yield-line method is a tool to analyze the load-carring capacity of a slab
11.21

Strip method

Introduction
• These drawbacks motivated Hillerborg in 1956 to develop a strip method for slab design
11.22

7
Strip method
Introduction
• This method is a lower-bound approach, based on satisfaction of equilibrium require-
ments everywhere in the slab
• According to the strip method a moment field is first determined that fulfills equilibrium
requirements, after that the reinforcement in the slab is designed for this moment field
• The strip method is a design method for direct calculation of the required reinforcement
• The strip method relies on the designer’s intuitive feel of the way the structure transmits
the load to the supports
11.23

2.2 Basic principles

Strip method
Governing equations
• The moment equilibrium equation is
∂ mx ∂ mxy ∂ my
+2 + = −q(x, y)
∂ x2 ∂ x∂ y ∂ y2
• According to the lower-bound theory, any combination of mx , my and mxy that satisfies the
equilibrium equation and meets boundary conditions is a valid solution
• The strip method assumes that the mxy = 0
• We can treat a slab as a system of twistless strips in x and y directions, each carryng a part
of the load q
∂ mx ∂ my
+ 2 = −q(x, y)
∂ x2 ∂y
∂ mx ∂ my
= −qx and 2 = −qy
∂ x2 ∂y
where qx + qy = q
11.24

2.3 Choise of load distribution

Strip method

Choice of load distribution

8
 • The load can be divided arbitrary between the x and y directions
• For a square slab the simplest load distribution is to split it equally in both directions
qx = qy = 2q
• The bending moments are
1 2
mx = my = qa
16
• This solution is not so practical or economical and requires great redistribution of moments
to achieve, together with excessive cracking and large deflections
11.25

Strip method

Choice of load distribution

• More complicated distribution of the load
• The slab is divided into strips with different width
• This design leads to a practical arrangement of reinforcement
• This solution is fully consistent with the equilibrium theory
11.26

9
Strip method

Example 1
• A reinforced concrete slab is loaded by a uniform load p
• Applying the strip method calculate the lower-bound value of the load
• Assumed main and secondary strips and load distribution
11.27

Strip method

10
Example 2
• A reinforced concrete slab is loaded by a uniform load p
• Applying the strip method calculate the lower-bound value of the load
• Assumed main and secondary strips and load distribution
11.28

Strip method

The End
• Imhotep- the 1-st engineer
• Any questions, opinions, discussions?
11.29

11