1

YIELD LINE THEORY AND ANALYSIS

1. In general, the various methods for the design and analysis of slabs apply only to those that are square or rectangular supported on two parallel sides or on all four

sides. The loads must also be uniformly distributed and there must not be large

openings on the panel. In practice, however, this is not always the case. There are

slabs that are triangular, circular, and trapezoidal; there are rectangular panels

supported on three sides only and sustain concentrated loads in addition to a

uniform load.

Yield line Analysis can be used under these conditions.

2. Yield Line Analysis of slabs is a method based on the formation of heavy cracks along lines dividing the slab into segments. The slab deforms and folds along these lines

immediately before failure. Each line, called yield line, is actually a series of plastic

hinges in a straight line formed between adjacent slab segments. As such, the yield

line serves as an axis of rotation for the segment.

3. It may be recalled that a plastic hinge along a beam is formed when the external moment at any section attains a magnitude equal to the flexural strength of that

same section. If the beam is simply supported, deflection in the plastic hinge

increases without any increase in the bending load until the structure collapses.

4. Thinking of a slab as a group of identical wide and shallow beams placed side by side to each other, the series of plastic hinges formed creates the yield line.

5. Consider for instance a slab with a uniform load and simply supported on two parallel edges. With the maximum moment occurring at midspan only, the yield line

is expected to occur there inasmuch as that is where moment is significant and

peaks.

6. In indeterminate structures, however, the formation of a single plastic hinge does not immediately cause collapse as the structure tends to redistribute the moments.

With moment redistribution, plastic hinges also form at different sections until the

structure totally fails.

7. Say a slab is fixed at both ends and carries a uniform load. The maximum negative moment occurs at the supports so the yield lines are expected to form there but the

slab does not fail because the moment is redistributed. As the load increases, a third

yield line forms at midspan where maximum positive moment occurs. It is only when

this third yield line is formed that the slab will collapse.

8. The foregoing concepts apply to slabs in general. Cracks are initially formed, then with increasing load, the cracks propagate gradually forming a yield line. Several

2

such yield lines may be formed as the bending moments are redistributed to

adjacent portions. Finally, the slab fails. This development of yield lines relative to

failure is called a mechanism.

9. Analysis by yield lines is an upperbound method. This is because the computed moment capacity of the slab is greater than the actual strength. Nevertheless, the

method is useful inasmuch as it may be resorted to in slabs that cannot be analyzed

by other methods due to unusual configuration. The use of a strength reduction

factor is therefore advantageous in this case.

10. Yield Line Analysis assumes the following :

The strength of the slab is controlled by flexure; The slab is ductile (the steel ratio is way below the balanced steel ratio)

enabling it to rotationally deform and redistribute the moments allowing

other yield lines to form and propagate until it finally fails;

The ratio of the span to the slab thickness (L/h) is within 10 to 40. Slabs within this range are called medium-thick slabs.

Patterns of Yield Lines and Notational Symbols

11. To use yield line analysis, the orientation and location of the yield line must first be approximated. The mechanism leading to failure must also be visualized.

The following are aids in drawing yield lines:

Lines must be straight; Lines must extend to the edges of the slab; Lines may intersect with other yield lines; Lines terminate at the point of intersection; Lines must pass through the intersection of the axes of rotation of adjacent slab

segments.

If the axes of rotations are parallel, the yield line must similarly be parallel. An axis of rotation passes through a column

As shown in the following figures, axes of rotations, in general, are found along lines

of support.

3

12. To identify supports, yield lines and axes of rotation in a slab, some of the following symbols are used in these notes:

Simple support

Fixed support

Free or unsupported edge

Axis of rotation

Yield line for positive moment

Yield line for negative moment

4

Yield Line Analysis Using the Principle of Virtual Work

13. In view of equilibrium, the sum external work done by the loads to cause a small arbitrary virtual deflection and the internal work done as the slab rotates at the yield

lines to accommodate the deflection, must be zero; the two are equal.

We, external work = Wi, internal work

14. The external work is equal to the product of the constant magnitude of the load and the distance through which the point of application of the load moves.

15. For instance, if we let the resultant load (or concentrated load as the case maybe) be R, and the virtual displacement is arbitrarily assigned a unit value, then:

external work, We = (R) (1)

16. If the load is not concentrated but instead distributed over an area or length, the external work is the product of the resultant and the displacement of the point of

application of its resultant.

17. The internal work, on the other hand, is the sum of the products of the yield moments per unit length (that is, the moment strength of the slab section normal to

the yield line), the length of the yield line, and the plastic rotation corresponding to

the virtual deflection.

18. The yield moment is the ultimate flexural strength of a section at the yield line. If mu is the ultimate flexural strength, l the length of the yield line, and the rotation,

then:

internal work, Wi = mu l

19. If the bar spacing or bar size in the slab change along a yield line, the flexural strength of the slab also varies. In this case the yield line may be divided into a

number of segments such that each segment has a constant slab section.

The internal work becomes:

Wi = (m1 l1 + m2 l2 + m3 l3 ++ mn ln )

Flexural Strength Along Yield lines

20. As noted earlier on, the moment used in calculating the internal work is the flexural strength of the section normal to the yield line. Frequently, however, yield lines

formed in a slab are not normal to the reinforcement. Furthermore, in a number of

5

slabs such as two way slabs, the steel bars run in two directions perpendicular to

each other (the slab is said to be orthotropic) with different effective depths. With

the yield lines skewed and reinforcements running in two directions, there will be

moment capacities in the orthogonal directions. In such as case, the flexural strength

along the yield line must be the moment with contributions from the two directions.

The moment is given by:

Mu = = = = mux cos

2 + + + + muy sin2

where: = = = = angle with of the yield line with respect the x-axis mu = flexural strength of the slab section normal to the yield line mux = flexural strength of the slab section normal to the x-axis muy = flexural strength of the slab section normal to the y-axis

21. If reinforcement is the same in both directions, slab reinforcement is said to be

isotropic, so that mux = muy . When this is the case, mu = mux = muy

6

Rotation of Yield Lines

22. The rotation of the yield line from its original horizontal position is evaluated by drawing a line from its initial position to the rotated position as viewed in a vertical

plane. Rotation is relative to the segments on both sides of the yield line.

Yield Line Patterns in Rectangular Slabs Supported on All Edges

23. The simplified yield line pattern in rectangular slabs supported on all edges is as shown. In reality however, the yield line forks as it nears the edge, forming corner

levers. The presence of the corners levers is generally neglected because

computations will be more tedious if this were considered. Besides, the difference in

calculations may not really be significant.

7

Yield Line in Triangular Slabs Supported on Two Edges

24. As shown, the yield line, inclined at an unknown angle, passes through the intersection of the axes of rotation and intercepts the free edge.

For instance, in the following figure, the yield line intercepting the free edge is

inclined with respect the horizontal at an angle . There exist a specific angle of

that yield line corresponding to the largest load the slab can safely sustain (the

smallest load at which the slab can fail) can be computed.

8

Recall that, depending on the kind of support, there may or may not be yield lines

parallel to the supports.

At very small slopes, assume = 1 at point O.

First note that the yield line intersects the free edge at an unknown point O at an

unknown angle. Then the green line drawn is perpendicular to the yield line at point

O and intersects the axes of rotation in order to determine the segment rotation .

9

The internal and external work can then be expressed in an equation having

unknown dimensions and/or angles expressed in terms of another as parameters.

One of the variables in the equation, say x, can be gradually increased so as to

determine the corresponding changes relative to the yield line and the load.

Yield Lines in Rectangular Slabs Supported on Three Edges

25. With three sides supported and the fourth edge unsupported, the free edge is intercepted by yield lines emanating from the corners. The lines are inclined at an

unknown angle as illustrated below.

10

As before, to determine rotations, a straight line is drawn perpendicular to the yield line

at the point where a virtual displacement of = 1 (point O in the figure) is imposed.

11

Yield Line Patterns Due To Concentrated Loads

26. When a slab fails due to the action of a concentrated load away from the corner or edge of the slab, negative yield lines form a nearly circular pattern while positive

yield lines radiate from the point of application of the concentrated load.

The collapse load is:

Pu = 2 ( mp + m

n)

Where:

Pu = Concentrated load at which the slab will fail

mp = positive resisting moment per unit length

mn = negative resisting moment per unit length

12

Illustrative Problems

Problem 1

The slab shown is considered to be simply supported on its two edges. Find the

maximum load Wu, the slab can carry using virtual work method.

fc=20 MPa

fy =415 MPa

Solution:

Being simply supported, only a positive yield line is formed located at midspan.

Note that there are two axes of rotation, one at each parallel support.

Also, since the axes of rotations are parallel, the yield line is parallel them.

13

As shown, the yield line divides the slab into two equal segment or plates of equal area.

Under its failure load, each side of the slab rotates about its end supports as a rigid

body.

Assume a unit value of deflection corresponding with very small slopes.

1=

8.1

== BA

8.1

1== BA radian

= total rotation

A B = +

9.0

1

8.1

12 =

= radian

Determine the internal work

Wi m = l

Since the yield line is perpendicular to the slab reinforcement, consider a unit strip of

slab also perpendicular to the yield line.

14

mmd

mmb

17030200

1000

===

( )( )( ) ( )

2

2

2

2

2

2

4

41000

(16)275 4

731.13

7310.0043

(1000)(170)

(1 0.59 )

(415)0.0043 1000 170 415 1 0.59(0.0043)

20

b

b

n

n

As n d

bAs d

s

As

As mm

As

bd

fyM bd fy

fc

M

=

=

=

=

=

= =

=

=

15

48857155.46 . /

48.86 . /n

n

M N mm meter

M KN m meter

==

0.90

43.97 . /nm M KN m m

== =

Notice that the reduced nominal strength is the flexural strength per meter of slab along

the yield line.

Thus,

iW m = l Where: m4.2=l

. 143.97 (2.4 )

0.9

117.25 .

i

i

KN mW m radian

m

W KN m

=

=

Compute the external work

We= sum of the products of the resultant loads on each segment and the displacement

of the point of action of the resultant

16

Using a unit displacement at the yield line, 1 =

At point where the resultant acts, displacement is 1

2 2

=

Work done by the two segments,

( )( ) ( )( )

ue

uue

WW

WWW

32.4

)2

1(4.28.1)

2

1(4.28.1

=

+=

Internal work = external work

KPaWm

KNW

W

W

WW

u

u

u

u

ei

14.27

14.27

32.4

25.117

32.425.117

2

=

=

=

==

17

Problem 2

Let the slab in problem 1 be fixed on one side and on a simply supported on the other

side. Negative bars are also provided on the fixed edge, as shown. Calculate the failure

load Wu.

Solution

Two parallel yield lines are formed- a negative yield line at the fixed support and a

positive yield line between supports.

Let the yield line between supports be located at an unknown distance x from the fixed edge.

Assume a virtual deflection of 1 unit at the yield line with corresponding rotations of A

and B

18

( )x

x

B

A

=

=

6.3

1

1

The total rotation @ C is:

( )xxcBAc

+=

+=

6.3

11

Find the internal work, Wi

i A A A C C CW m m = +l l

Consider a meter strip of slab, so that b=1 m =1000 mm

19

Unit strip with bars perpendicular to the negative yield line at the fixed support

( )

( )( )

2

2

2

2

(1 0.59 )

100016

200 4

1005.31

1005.310.006

1000 170

4150.90(0.006)(1000)(170) (415) 1 0.59(0.006)

20

A

A

fym bd fy

fc

As

mm

m

=

=

=

= =

=

60007594.28 .

60.01 . /A

A

m N mm

m KN m m

==

Determine mC Unit strip with bars perpendicular to the positive yield line between supports

20

( )

( )( )

2

2

2

(1 0.59 )

100016

275 4

731.13

731.130.0043

1000 170

C

fym bd fy

fc

As

As mm

=

=

=

= =

( )( )( ) ( ) ( )2 4150.90 0.0043 1000 170 415 1 0.59 0.004320

4.3971439.91 .

43.97 . /

C

C

C

m

m N mm

m KN m m

= ==

( ) ( ) ( ) ( )

( )

xxW

xxxW

xxxW

i

i

i

+=

++=

++=

6.3

528.10555.249

6.3

528.5.10528.105024.144

4.2)6.3

11(97.434.2)

1(01.60

Compute external work

21

( )( ) ( )( )( )xWXWW

xWxWW

uue

uue

+=

+=

6.32.12.12

14.26.3

2

14.2

( )

( )

( ) ( )16.3428.24766.57

6.3

428.24766.5732.4

6.3

528.10555.249

32.46.3

264.117296.277

32.4

+=

=

+

=+

=

==

xxW

Wxx

Wxx

Wxx

WW

WW

u

u

u

u

ei

ue

To obtain the critical value of Wu, find the value of x that minimizes Wu.

Find the first derivative of Wu with respect to x then set to zero.

( ) ( )( )

( )( ) ( )( )

( )

( )( ) ( )

22

222

22

22

22

428.24766.5792.41565.748

428.24428.246.3766.57

6.3

428.24766.57

6.3

428.24766.57

6.3

10428.2406.31766.570

xxxO

xxxO

xxO

xxdx

dW

x

x

x

x

dx

dW

u

u

++=++=

+=

+=

+=

( ) ( ) ( )( )( )

( )

( ) 181.2

,294.10676.66

473.27092.415

338.332

65.748338.33492.41592.415

065.74892.415338.33

338.3392.41565.748

2

2

2

=

=

+=

+=

=++=

+

x

discardx

x

x

xx

xxO

22

mx 181.2=

Substitute in equation 1

( )181.26.3428.24

181.2

766.57

+=UW

KPaWM

KNW

U

U

70.43

70.432

=

=

Note:

In this problem, it so happened that the resulting equation is easy to solve.

For other situations, successive approximations may be done to obtain the solution.

Problem 3

A simply supported two-way slab is reinforced in the short direction with in mm12 bars spaced @ 125 mm o.c., and mm12 bars @ 150mm o.c. in the long direction. Determine the magnitude of the uniformly distributed failure load producing the yield

line pattern shown.

MPacf 20= MPafy 275=

23

NOTE: This given pattern may not be the critical pattern at failure.

Solution:

Find the flexural strengths in the X and Y directions

Consider a unit strip of slab in both directions

24

SHORT DIRECTION (4 m span; bars parallel to the y-axis)

( )

( )( ) 007.0124100078.904

78.904124125

1000 22

==

==

mmAs

NOTE: Bars in the y-direction resist bending with respect the x-axis and therefore

provide a moment resistance mx per unit length

2 1 0.59X Sfy

m bd fyfc

=

( ) ( )( ) ( ) ( ) ( )2 2750.007 1000 124 275 1 0.59 0.00720

25126162.34 .

25.13 . /

x

X

X

m

m N mm

m KN m m

=

==

LONG DIRECTION (6 m span; bars parallel to the x-axis)

25

( )

( )( ) 0067.0112100098.753

98.753124150

1000 22

==

==

mmAs

NOTE: Bars in the x-direction resist bending with respect the y-axis and therefore

provide a moment resistance my per unit length

( )( )( ) ( ) ( ) ( )2 2750.90 0.0067 1000 112 275 1 0.59 0.006720

19670470.87 .

19.67 . /

y

y

y

m

m N mm

m KN m m

=

=

=

Considering the inclined yield line, the moment normal to the line is: 2 2cos sinX ym m m = +

( ) ( )2 0 2 025.13cos 45 19.67sin 4522.40 .

m

m KN m

= +

=

Determine yield line rotations

26

As before, assume a virtual deflection of 1= at the center of the slab at point f and along c as shown. Obviously, in this case, a = b = d. Moreover, note that d is drawn

simply as an extension of a in order to determine the rotation.

Rotation of inclined yield line b (due to symmetry, the rotation is the same for all inclined yield lines)

Where: 22== da

22

1122

11

==

==

d

a

df

af

ad = total rotation of inclined yield line

27

radianad

ad

ad

ad

ad

ad

707.02

2

2

2

2

1

2

122

2

2

1122

1

22

1

=

=

=

=

=+=

+=

Since all inclined lines have the same rotations,

let;

707.0=i radian for each inclined line

Rotation of horizontal yield line, h

1 1

2 21

h L R

h

h

= +

= +

=

Compute internal work done

i

i i i i X h h

W m

W m m

= = +

l

l l

28

Where:

=il Total length of inclined yield lines

h =l Length of horizontal yield lines

( )( )( ) ( )( )( )

8

2

4 22.40 8 0.707 25.13 2 1

229.374 .

i

i

h

i

i

m m

m

W

W KN m

=

==

= +

=

l

l

Find external work done

Let:

Wu= magnitude uniform load over the slab in each segment. The resultant of Wu, passes

through the centroid of the corresponding segment, as shown in the following

figure identified as points 1,2,3,4,5,6,7 and 8. The corresponding

deviations/deflections of those points are 1, 2, 3, 4, 5, 6, 7, and 8, with

resultants R1, R2, R3, R4, R5, R6, R7, and R8, respectively.

Resultant ( )UW= (area)

29

( )( )

( )

( )( )

( )( )

( )( )[ ]( )

5.0

4

5.012

1

4223

1

4222

13

1

3

1

4242

13

11

3

1

3

1

2222

1

5

5

4

4

3

3

2

2

1

1

==

==

==

=

=

=

==

==

===

=

=

U

UU

UU

UU

UU

WR

WWR

WWR

WWR

WWR

( ) ( )

Ue

UU

UUUUUUe

e

U

U

U

WW

WW

WWWWWWW

RRRRRRRRW

WR

WR

WR

33.9

3

12

3

14

3

125.045.04

3

12

3

14

3

12

3

1

23

1

43

1

2

8877665544332211

8

8

7

7

6

6

=

+

+

+++

+

+

=

+++++++=

=

=

=

=

=

=

30

KPaWm

KNW

W

WW

U

U

U

ei

56.24

56.24

33.9374.229

2

=

=

==