Analysis and Design of Reinforced Concrete Structures 2: Dr. Hasan Albegmprli

Northern Technical University
Engineering Technical College

Building and Construction Engineering Department
Analysis and Design of Reinforced Concrete Structures 2
Dr. Hasan ALBEGMPRLI

It is an application for non-rectangular supports distances,
for irregular shape as a triangular and circular slab and for
slabs supported on three sides or less and free for others.
Also it is applicable for partial loading or point loading.
The method can used to analysis the effect of opening in
slab system.
• Yield line: Is a crack in a reinforced concrete slab occurs
which the reinforcing bars have yielded and along which
plastic hinge rotation.
• Yield line theory: Is an ultimate load analysis. It
establishes either the moment at the point of failure or the
load at which an element will fail.
• Yield lines pattern: yield lines divide the slab into
individual regions which pivot about the axes of rotation.
• The figure below shows one way simply
supported slab with uniform load w kN/m
upon yielding, the curvature of the slab at the
yielding section increases sharply, and
deflection increases disproportionality.
• The elastic curvatures along the slab span
are small compared with the curvature
resulting from plastic deformation at the yield

line.
• Therefore it is assumed that slab segments
between yield lines and supports remain rigid.

• Consider now the fixed – fixed slab shown in
figure which is indeterminate structure. If the
slab is equally reinforced for (+ ve) and (-ve)

moments, yield lines form at locations of
maximum moment (supports), after increasing
the load a third yield line form at mid span and

the slab is now a mechanism.
• From above it is point out that moment ratios at
the collapse stage depend upon the
reinforcement provide, not upon the results of
elastic analysis.
• A yield line is called positive if it is associated

with compression at top and tension at bottom

and vice versa.
Rules for yield lines
1. Yield lines are straight lines because they represent the intersection of two planes.
2. Yield lines represent axes of rotation.
3. The supported edge of the slab will also establish axes of rotation. If the edge is
fixed, a negative yield line may form providing constant resistance to rotation .If
the edge is simply supported, the axis of rotation provides zero restraint.
4. An axis of rotation will pass over any column support. Its
orientation depends on other considerations.
5. Yield line form under concentrated loads, radiating outward
form the point of applications.
6. A yield line between two slab segments must pass through the
intersection of the axes of rotation of the adjacent slab segment.

Upper and lower bound theorems
•If the solution to a problem gives load less than the true carrying capacity,
it is called lower bound solution.
•If the solution to a problem gives load more than the true carrying
capacity, it is called upper bound solution.
•The yield line method of analysis for slabs in an upper bound method.
• The yield line phenomenon involves:
– a slab under increasing loads where cracking and reinforcement yielding
occur in the most highly stressed zone (i.e. around maximum moment)
– the highly stressed zone normally acts as a plastic hinge where the subsequent
loads are distributed to other region of the slab

– cracks develop forming patterns of yield lines until a mechanism is formed,
– collapse is then indicated by increasing deflection under constant load

• Deformation of slab with yield lines
Notati
ons
If the reinforcement is the same in both directions, then the moment capacity is the same and the slab is called isotropic.
Typical yield line patterns
Method of yield line analysis
There are two methods of yield line analysis of slabs:

– Equilibrium method
– Virtual work method
• In either method, a yield line pattern is assumed so that a collapse
mechanism is produced. Then for that failure mechanism, the
geometric parameters that define the exact location and orientation
of the yield lines are determined and also the relation between
applied loads and resisting moments is solved.
• It is necessary to investigate all possible mechanisms for any slab
to confirm that the correct solution, giving the lowest failure load,
has been found. For example the following rectangular slab (SNS)
may fail by either of the two mechanisms shown.
Alterative mechanisms for a slab supported on three sides
The Equilibrium method
• In this method each segment studied as a free body, must be in
equilibrium under the action of loads, resisting moments along the
yield lines and the reactions and shear along support lines.
• At the yield line, the twisting moment is zero. The shear is also
zero at most of yield lines.

Example:
A square slab is simply supported along all sides and is to be

isotopically reinforced. Determine the resisting moment mu = φ mn
per linear meter required just to sustain a uniformly distributed
factored load of q kN/m2.
Solution:
• Conditions of symmetry indicate the yield line pattern

shown considering the equilibrium of segment A.
• The resisting moment along each inclined edge of

segment A is
𝑚𝐿 𝑚𝐿 1 𝑚𝐿
The x-component of moment along inclined edge = cos 45 = 𝑥( ) =
2
ξ2 ξ2 ξ2
Σ M along a - b = 0
1 𝐿 1 𝐿 𝑚𝐿 𝑚𝐿
𝑞൬𝑥𝐿𝑥 ൰൬𝑥 ൰− − =0
2 2 3 2 2 2
𝑞𝐿2
𝑚=
24
If mx represents the yield moment about the x-axis and my about the y-axis, the yield moment about
an axis at angle α with the x-axis will be
mα = mx cos2 α + my cos2(90- α)
= mx cos2 α + my sin2 α
When mx = my = m this leads to mα = m.
Example:
A uniformly loaded, one way continuous slab has 3 m

span and is required to provide a resistance to +ve
bending φmn = 22 kN.m through the span. –ve moment
capacity of 22 kN.m/m at A and 33 kN.m/m at c.
determine the load capacity of the slab
Solution:
Tacking strip of 1m width
Σ MA = 0
Σ MC = 0
Solving eqs.1 and 2 for x and q
x = 1.416 m
q = 43.889 kN/m2

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Northern Technical University

Engineering Technical College

Analysis and Design of Reinforced Concrete Structures 2

Dr. Hasan ALBEGMPRLI

for irregular shape as a triangular and circular slab and for

slabs supported on three sides or less and free for others.

Also it is applicable for partial loading or point loading.

The method can used to analysis the effect of opening in

resulting from plastic deformation at the yield

slab is equally reinforced for (+ ve) and (-ve)

the load a third yield line form at mid span and

with compression at top and tension at bottom

2. Yield lines represent axes of rotation.

orientation depends on other considerations.

5. Yield line form under concentrated loads, radiating outward

form the point of applications.

intersection of the axes of rotation of the adjacent slab segment.

it is called lower bound solution.

capacity, it is called upper bound solution.

– a slab under increasing loads where cracking and reinforcement yielding

loads are distributed to other region of the slab

– collapse is then indicated by increasing deflection under constant load

There are two methods of yield line analysis of slabs:

• In this method each segment studied as a free body, must be in

equilibrium under the action of loads, resisting moments along the

zero at most of yield lines.

A square slab is simply supported along all sides and is to be

• Conditions of symmetry indicate the yield line pattern

• The resisting moment along each inclined edge of

an axis at angle α with the x-axis will be

A uniformly loaded, one way continuous slab has 3 m

Solving eqs.1 and 2 for x and q

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