Chapter 1 – Yield Line Analysis for slabs

SD:Chapter 1 1

Chapter 1 – Yield Line Analysis for slabs

Dr. -Ing. Adil Z.Dr.-Ing. Girma Z.

SD:Chapter 1 2

Behavior of slabs loaded to failure in flexure

• Four or more stages:i. Before cracking the slab acts as an elastic plate, and

for short time loads, the deformations, stresses and strains can be predicted from an elastic analysis.

ii. After cracking and before yielding of the reinforcement, the slab no longer has a constant stiffness, because the cracked regions have a lower flexural stiffness, EI than the uncracked regions and the slab is no longer isotropic because the crack pattern may differ in the two directions.

SD:Chapter 1 3


– Although these conditions violate the assumptions in elastic theory, tests indicate that the elastic theory still predicts the moments adequately. Generally normal building slabs are partially cracked under service loads.

iii. yielding of reinforcement eventually starts in one or more region of high moment and spreads through the slab as the moments are redistributed from yielded regions to areas that are still elastic. The progression of yielding through a slab fixed on four edges is illustrated in Figure (next slide)

SD:Chapter 1 4


SD:Chapter 1 5


– With further load, the regions of yielding known as yield lines, divide the slab into a series of trapezoidal and triangular elastic plates as shown in Figure (d) above. The loads corresponding to this stage of loading can be estimated by using yield-line analysis (plastic method analysis)

iv. Although the yield lines divide the slab to form a mechanism, the hinges jam with increased deformation, and the slab forms a very flat compression arch as shown in Figure (next slide)(avail stiff support). This stage of loading usually is not considered in design

SD:Chapter 1 6


SD:Chapter 1 7


• Yield line method is a plastic method of analysis of slabs

• A yield line analysis uses rigid plastic theory to compute the failure loads corresponding to given plastic moment resistance in various parts of the slab

• A yield line method is an upper bound method whereas the strip method is lower bound method.

SD:Chapter 1 8


• Upper bound theorem: If, for a small increment of displacement, the internal work done by the slab, assuming that the moment at every plastic hinge is equal to the yield moment and that boundary conditions are satisfied, is equal to the external work done by the given load for that same small increment of displacement, then that load is an upper bound of true carrying capacity.

SD:Chapter 1 9


• Lower bound theorem: If, for a given external load, it is possible to find a distribution of moments that satisfies equilibrium requirements, with the moment not exceeding the yield moment at any location, and if the boundary conditions are satisfied, then the given load is a lower bound of the true carrying capacity.

SD:Chapter 1 10


• If the lower bound conditions are satisfied, the slab can certainly carry the given load, although a higher load may be carried if internal distributions of moment occur. If the upper bound conditions are satisfied, a load greater than the given load will certainly cause failure, although a lower load may produce collapse if the selected failure mechanism is incorrect in any sense.

SD:Chapter 1 11


• 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

SD:Chapter 1 12


• Deformation of slab with yield lines

SD:Chapter 1 13


Guidelines for establishing axes of rotation and YL1. YL are straight lines because they represent

intersection of two planes2. YL represent axes of rotation3. The supported edges of the slab are axes of rotation.

If the edge is fixed negative YL 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

SD:Chapter 1 14


Guidelines for establishing axes of rotation and YL5. YL form under concentrated loads, radiating

outward from the point of application.6. A YL between two slab segments must pass through

the point of intersection of the axes of rotation of the adjacent slab segments.

SD:Chapter 1 15


SD:Chapter 1 16


SD:Chapter 1 17


SD:Chapter 1 18


• Method of yield line analysis• There are two methods of yield line analysis of slabs:

– the equilibrium method– the virtual work method

• In either method, a YL 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.

SD:Chapter 1 19


• 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.

SD:Chapter 1 20

Alterative mechanisms for a slab supported on three sides



• Equilibrium method of Analysis• It makes use of the equilibrium equations for

individual segment to obtain the collapse load. • The FBD represented by each collapsing

segment is in equilibrium under– applied loads,– yield moments and– reactions or shears along support lines.

SD:Chapter 1 21

SD:Chapter 1 22


• For demonstration purpose consider the one way slab uniformly loaded and is continuous as shown in Fig. below.

• Let the slab with span L is reinforced to provide resistance of m2 kN.m per m through the span and m1 and m3 kN.m per m at the two supports.

• Suppose it is desired to determine the collapse load wu.

SD:Chapter 1 23

A C B

w u

X L -X

L

w u w u

m 3

m 2

m 2m 1

X L -X

(a)

(b )

(c)


SD:Chapter 1 24


• For a known yield moments m1, m2, m3, a trial location of the positive yield line is assumed.

• Normally for a given loading and correct collapse pattern the solution is unique. However if a different pattern is assumed, this solution can describe nothing which pattern is the governing one. Hence, it becomes essential to use the energy approach in completed problem for further verification.

SD:Chapter 1 25


• For the problem posed, consider the FBD in Figure (c);

• From left segment:

• From right segment:• Solving for wu from the two expressions and

equating, one may obtain a practical solution for x as:

02 02

21 xwmmM uA

02)( 0

2

32 xLwmmM uB

21

13

13

21 11mmmmL

mmmmx

SD:Chapter 1 26


• For instance, when L = 3m, m1 = m2 = 7kNm/m and m3 = 10KN.m/m x = 1.427m and wu = 13.75 KN/m2

• When m1 = m3 = 10kN.m/m and m2 = 7kNm/m,

L = 3mx = 1.5m and wu = 15.11 KN/m2

SD:Chapter 1 27


• With the equilibrium method considerable care must be taken to show all forces acting on each element, including twisting moments, especially when several YLs intersect or when YLs intersect free edges.

• At such locations, off-setting vertical nodal forces are required at the intersection of YLS.

• Because of the possibility of that the nodal forces will be given the wrong sign or location, some building codes don’t recommend the methods.

SD:Chapter 1 28


• Virtual work method of Analysis• Based on principle that work done by external

forces in undergoing a small virtual displacements is equal to the internal virtual work done in rotations along yield lines, the ultimate load which the slab can sustain is determined.

WE = WI

SD:Chapter 1 29


• To develop suitable expression for each work, let wu be the uniform distributed load,

• • where: – xy = virtual displacement at load point considered–Wu = resultant of the load on each segment– ∆ = the corresponding displacement at centroid of

the load in each segment

uxyuE WdxdywW

SD:Chapter 1 30


• Since the relative rotation of the surface takes place about yield line, one may obtain

• Where: – mun = the ultimate moment across any yield line.– Ly = length of yield line– θn = relative rotation of the two adjacent plates

perpendicular to the yield line

ynunI LmW

SD:Chapter 1 31


• For demonstration purpose, consider the one-way slab of the previous example.

w u

X L -X

L

w u w u

m 3

m 2

m 2m 1

X L -X

21

SD:Chapter 1 32


• The slab is reinforced at left and right supports, and in span in such a way that the capacities of the respective sections are m1, m3 and m2.

• Let ∆ be the plastic deflection at x-distance from left hand support to positive yield line

SD:Chapter 1 33


• External work

• Internal work

• For small deflection

strip mper 21

2)(

2LwxLwwxW uE

strip mper )()( 232121 mmmmLmW yI

xLx

21 and

xLmm

xmmWI

)()( 3221

SD:Chapter 1 34


• By principle of virtual work; WE =WI which simplifies to give:

• To obtain the minimum collapse load

)(

)()(2 3121

xLxLmmxmmLwu

0dx

dwu

0)2()()()()( 2312131 xLLmmxmmLxLxLmm

SD:Chapter 1 35


• This gives the practical values of x such that:

• Thus, identical solution to the equilibrium methods is obtained

21

13

13

21 11mmmm

Lmmmm

x

SD:Chapter 1 36


• Moments along Skewed yield lines

Consider a two way slab orthogonally reinforced where the yield lines are inclined at an angle α with one of the principal axes.

SD:Chapter 1 37


• The Fig. shows an orthogonal gird of reinforcement having moment resistance my and mx per unit length about the y axis and x axis, respectively.

• the resisting moment per unit length along the α axis provided by the y direction bars is:

• the resisting moment per unit length perpendicular to α axis provided by the y direction bars is:

2cos

coscos

xx

y mu

umm

sincoscossin

xx

y mu

umm

SD:Chapter 1 38


• the resisting moment per unit length along the α axis provided by the x direction bars is:

• the resisting moment per unit length perpendicular to α axis provided by the x direction bars is:

2sin

sinsin

yy

x mv

vmm

sincossincos

yy

x mv

vmm

SD:Chapter 1 39


• Thus for the combined set of bars:• the resisting normal moment per unit length

measured along the α axis is: …(1)

• the resisting torsional moment per unit length measured along the α axis is:

…(2)

22 sincos yx mmm

sincossincos yx mmm

SD:Chapter 1 40


• An isotropically reinforced slab is one in which the ultimate moment per unit length of the slab is the same in two orthogonal directions– If a slab is isothropically reinforced with mx = my =

m, equations (1) and (2) become:

– Therefore the ultimate moment resistance in an isotropically reinforced slabs in any direction is the same.

0 mandmm

SD:Chapter 1 41


• An orthotropically reinforced slab is one in which the ultimate moment per unit length of the slab is different in the two orthogonal directions– If a slab is orthothropically reinforced with mx = m

and my = μm, equations (1) and (2) become:

)sin(cos 22 mm

cossin)1( mm

SD:Chapter 1 42


• Effects of restrained corners• Corner lever is the effect of forking of the yield

line before reaching the corner. To this effect the following schematic sketches the resulting yield pattern at the corners.

SD:Chapter 1 43

U pli f t when the cor ner s ar e notheld or f astened

Cor ner s held down but not wellr ei nf or ced f or negati ve moment

H oggi ng yi eld li ne


SD:Chapter 1 44Development of corner levers in a simply supported slab


SD:Chapter 1 45

Chapter 1 – Yield Line Analysis for slabsm’=0

m’= m

m’= m/4

m’= m/2

SD:Chapter 1 46


• The triangular segment fails to form when the negative reinforcement is large and hence, the simple diagonal yield line in to the corner is correct with out modification.

SD:Chapter 1 47


• Slabs with more than one variable• So far the slabs consider have only one

variable dimension which defines the yield line mechanism. When the slab has more than one variable, the work equation, together with equations obtained by differentiating with respect to each unknown, give the necessary expressions to obtain solution. This can be illustrated using the following example.

SD:Chapter 1 48

L

L

(1-1)L

1L

2L (1 -2 2)L 2L


SD:Chapter 1 49


• In this case first develop suitable expression from the work relation for wu in terms of 1 & 2. Then

• and

• Will provide two additional equations to make the problem solvable.

01

uw 02

uw

SD:Chapter 1 50


Advantages of yield line theory• It is a powerful analytical tool.• It can be used for slabs of any shape,

supported in a variety of ways.• Concentrated, UD and partially distributed

loads can be considered.• The effects of holes of any size can be

included.

SD:Chapter 1 51


Limitations of yield line theory• If the selected failure mechanism is not the controlling

one or the location of the YL are not exactly correct, it may predict a collapse load greater than the true collapse load. Unsafe !!

• The slab section has to be checked for the adequateness of rotation capacity.

• It is presumed that premature failure will not occur due to shear or torsion and the cracking and deflections at service loads will not be excessive.

SD:Chapter 1 52


• References– Nilson, Darwin and Doran, “Design of Concrete

Structures”, 14th Edition, 2010, Chapter 14.– MacGregor and Wight, “Reinforced Concrete,

Mechanics and Design”, 6th Edition, 2012, Chapter 14.

Chapter 1 – Yield Line Analysis for slabs

Documents

Transcript of Chapter 1 – Yield Line Analysis for slabs