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, however, this is not always always the case. There are 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 nalysis can be used under these conditions. !. Yield "ine nalysis of slabs is a method based on the formation of heavy crac#s along lines dividing the slab into segments. The slab deforms and folds along these lines immediately before failure. $ach line, called yield line, is actually a series of plastic hinges in a straight line formed between ad%acent slab segments. s such, the yield line serves as an a&is of rotation for the segment. '. It may be recalled that a plastic hinge along a beam is formed when the e&ternal moment at any section attains a magnitude equal to the fle&ural 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. (. Thin#ing 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. ). *onsider for instance a slab with a uniform load and simply supported on two parallel edges. +ith the ma&imum moment occurring at midspan only, the yield line is e&pected to occur there inasmuch as that is where moment is significant and pea#s. . In indeterminate structures, however, the formation of a single plastic hinge does not immediately cause collapse as the structure tends to redistribute the moments. +ith moment redistribution, plastic hinges also form at different sections until the structure totally fails. -. ay a slab is fi&ed at both both ends and carries carries a uniform load. The ma&imum negative moment occurs at the supports so the yield lines are e&pected to form there but the slab does not fail because the moment is redistributed. s the load increases, a third yield line forms at midspan where ma&imum positive moment occurs. It is only when this third yield line is formed that the slab will collapse. /. The foregoing concepts apply to slabs in general. *rac#s are initially formed, then with increasing load, the crac#s propagate propagate gradually forming forming a yield line. everal everal 1
such yield lines may be formed as the bending moments are redistributed to ad%acent portions. 0inally, the slab fails. This development of yield lines relative to failure is called a mechanism. . nalysis by yield lines is an upper2bound method. This is because the computed moment capacity capacity of the slab is greater greater than the actual strength. 3evertheless, 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. 14. Yield "ine nalysis assumes the following 5 • •
•
The strength of the slab is controlled by fle&ure; The slab is ductile 6the steel ratio is way below the balanced steel ratio7 enabling it to rotationally deform and redistribute the moments allowing other yield lines to form and propagate until it finally fails; fails; The ratio of the span to the slab thic#ness 6"8h7 is within 14 to (4. labs within this range are called medium-thick slabs. 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 appro&imated. The mechanism leading to failure must also be visualized. The following are aids in drawing yield lines5 • • • • • • •
"ines must be straight; "ines must e&tend to the edges of the slab; "ines may intersect with other yield lines; "ines terminate at the point of intersection; i ntersection; "ines must pass through the intersection of the a&es of rotation of ad%acent slab segments. If the a&es of rotations are parallel, the yield line must similarly be parallel. n a&is of rotation passes through a column
s shown in the following figures, a&es of rotations, in general, are found along lines of support.
2
such yield lines may be formed as the bending moments are redistributed to ad%acent portions. 0inally, the slab fails. This development of yield lines relative to failure is called a mechanism. . nalysis by yield lines is an upper2bound method. This is because the computed moment capacity capacity of the slab is greater greater than the actual strength. 3evertheless, 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. 14. Yield "ine nalysis assumes the following 5 • •
•
The strength of the slab is controlled by fle&ure; The slab is ductile 6the steel ratio is way below the balanced steel ratio7 enabling it to rotationally deform and redistribute the moments allowing other yield lines to form and propagate until it finally fails; fails; The ratio of the span to the slab thic#ness 6"8h7 is within 14 to (4. labs within this range are called medium-thick slabs. 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 appro&imated. The mechanism leading to failure must also be visualized. The following are aids in drawing yield lines5 • • • • • • •
"ines must be straight; "ines must e&tend to the edges of the slab; "ines may intersect with other yield lines; "ines terminate at the point of intersection; i ntersection; "ines must pass through the intersection of the a&es of rotation of ad%acent slab segments. If the a&es of rotations are parallel, the yield line must similarly be parallel. n a&is of rotation passes through a column
s shown in the following figures, a&es of rotations, in general, are found along lines of support.
2
1!. To identify supports, yield lines and a&es a&es of rotation in a slab, some of the following symbols are used in these notes5 •
imple support
•
0i&ed support
•
0ree or unsupported edge
•
&is of rotation
•
Yield line for positive moment
•
Yield line for negative moment
3
Yield Line Analysis Using the Prini!le of "irt#al $or%
1'. In view of equilibrium, the sum e&ternal wor# done by the loads to cause a small arbitrary virtual deflection and the internal wor# done as the slab rotates at the yield lines to accommodate the deflection, must be zero; the two are equal. $e& e'ternal (or% ) $ i& internal (or%
1(. The e'ternal (or% 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. 1). 0or instance, if we let the resultant load 6or concentrated load as the case maybe7 be R, and the virtual displacement is arbitrarily assigned a #nit *al#e, then5 e'ternal (or%& $e ) +R, +-,
1. If the load is not concentrated but instead distributed over an area or length, the e&ternal wor# is the product of the resultant and the displacement of the point of application of its resultant. 1-. The internal (or%, on the other hand, is the sum of the products of the yield moments per unit length 6that is, the moment strength of the slab section normal to the yield line7, the length of the yield line, and the plastic rotation corresponding to the virtual deflection. 1/. The yield moment is the ultimate fle&ural strength of a section at the yield line. If m# is the ultimate fle&ural strength, l the length of the yield line, and 9 the rotation, then5 internal (or%& $i ) m# l . 1. If the bar spacing or bar size in the slab change along a yield line, the fle&ural 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 wor# becomes5 $i ) +m- l-
/ m0 l0 / m1 l1 /2/ mn ln , .
3le'#ral Strength Along Yield lines
!4. s noted earlier on, the moment used in calculating the internal wor# is the fle&ural strength of the section normal to the yield line. 0requently, however, yield lines formed in a slab are not normal to the reinforcement. 0urthermore, in a number of
4
slabs such as two way slabs, the steel bars run in two directions perpendicular to each other 6the slab is said to be orthotro!i7 with different effective depths. +ith the yield lines s#ewed and reinforcements running in two directions, there will be moment capacities in the orthogonal directions. In such as case, the fle&ural strength along the yield line must be the moment with contributions from the two directions. The moment is given by5 2 2 Muα = = mux cos α α + muy sin α
where: α
=
muα mux muy
angle with of the yield line with respect the x-axis = flexural strength of the slab section normal to the yield line = flexural strength of the slab section normal to the x-axis = flexural strength of the slab section normal to the y-axis
!1. If reinforcement is the same in both directions, slab reinforcement is said to be isotro!i, so that
mu& : muy . +hen this is the case, muα : mu& : muy
5
Rotation of Yield Lines
!!. 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. otation is relative to the segments on both sides of the yield line.
Yield Line Patterns in Retang#lar Slabs S#!!orted on All Edges
!'. The simplified yield line pattern in rectangular slabs supported on all edges is as shown. In reality however, the yield line for#s 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.
6
Yield Line in Triang#lar Slabs S#!!orted on T(o Edges
!(. s shown, the yield line, inclined at an un#nown angle, passes through the intersection of the a&es of rotation and intercepts the free edge.
0or instance, in the following figure, the yield line intercepting the free edge is inclined with respect the horizontal at an angle = . There e&ist a specific angle of that yield line corresponding to the largest load the slab can safely sustain 6the smallest load at which the slab can fail7 can be computed.
7
ecall that, depending on the #ind of support, there may or may not be yield lines parallel to the supports.
t very small slopes, assume > : 1 at point ?. 0irst note that the yield li ne intersects the free edge at an un#nown point ? at an un#nown angle. Then the green line drawn is perpendicular to the yield line at point ? and intersects the a&es of rotation in order to determine the segment rotation 9.
8
The internal and e&ternal wor# can then be e&pressed in an equation having un#nown dimensions and8or angles e&pressed in terms of another as parameters. ?ne of the variables in the equation, say &, can be gradually increased so as to determine the corresponding changes relative to the yield line and the load. Yield Lines in Retang#lar Slabs S#!!orted on Three Edges
!). +ith 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 un#nown angle as illustrated below.
9
s before, to determine rotations, a straight line is drawn perpendicular to the yield line at the point where a virtual displacement of > : 1 6point O in the figure7 is imposed.
10
Yield Line Patterns D#e To 4onentrated Loads
!. +hen 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 is5 !
n
P# ) 0 5 + m / m ,
+here5 P# : *oncentrated load at which the slab will fail ! m : positive resisting moment per unit length n m : negative resisting moment per unit length
11
Ill#strati*e Problems Problem The slab shown is considered to be simply supported on its two edges. 0ind the ma&imum load W u, the slab can carry using virtual wor# method. @ fc :!4 ABa fy :(1) ABa
olution5
12
s shown, the yield line divides the slab into two equal segment or plates of equal area. Cnder its failure load, each side of the slab rotates about its end supports as a rigid body. ssume a unit value of deflection corresponding with very small slopes. ∆ =1
θ A = θ B = θ A = θ B =
∆
1.8 1 1.8
radian
θ = total rotation
θ = θ A + θ B
1 1 radian = 1.8 0.9
θ = 2
Determine the internal wor#
Wi = mlθ
ince the yield line is perpendicular to the slab reinforcement, consider a unit strip of slab also perpendicular to the yield line.
13
b = 1000mm d = 200 − 30 = 170mm As = n As = As =
π 4
b π
d b
2
d b 2
s 4 1000 π 275 4
(16)
As = 731.13mm
ρ = ρ =
2
2
As bd 731 (1000)(170) 2
= 0.0043
M n = ρ bd fy (1 − 0.59 ρ
fy fc
)
M n = ( 0.0043)(1000)(170)
2
( 415) 1− 0.59(0.0043)
(415) 20
14
M n = 48857155.46N .mm / meter M n = 48.86KN .m / meter
φ = 0.90 m = φ M n = 43.97 KN .m / m
3otice that the reduced nominal strength is the fle&ural strength per meter of slab along the yield line.
Thus, Wi = mlθ +here5
l=
Wi = 43.97
2.4m
KN .m m
1 (2.4m ) 0.9 radian
Wi = 117.25KN .m
*ompute the e&ternal wor# +e: sum of the products of the resultant loads on each segment and the displacement of the point of action of the resultant
15
Csing a unit displacement at the yield line, ∆ = 1 t point where the resultant acts, displacement is
∆
2
=
1 2
+or# done by the two segments, 1 1 W e = (1.8)(2.4 )W u ( ) + (1.8)(2.4 )W u ( ) 2 2 W e = 4.32W u
Internal wor# : e&ternal wor# W i = W e 117.25 = 4.32W u
W u =
117.25 4.32
W u = 27.14
KN 2
m W u = 27.14 KPa
16
Problem 0 "et the slab in problem 1 be fi&ed on one side and on a simply supported on the other side. 3egative bars are also provided on the fi&ed edge, as shown. *alculate the failure load +u.
olution Two parallel yield lines are formedE a negative yield line at the fi&ed support and a positive yield line between supports. "et the yield line between supports be located at an un#nown distance & from the fi&ed edge. ssume a virtual deflection of 1 unit at the yield line with corresponding rotations of 9 and 9<
17
θ A = θ B =
1
x 1
(3.6 − x )
The total rotation F * is5
θ c = θ A + θ B θ c =
1
x
+
1
(3.6 − x )
0ind the internal wor#, +i Wi = m Aθ Al A + mCθ C l C
*onsider a meter strip of slab, so that b:1 m :1444 mm
18
Cnit strip with bars perpendicular to the negative yield line at the fi&ed support
fy
2
m A = φρ bd fy (1 − 0.59 ρ As =
1000 π 200 4
(16 )
= 1005.31mm
ρ =
fc′
)
2
2
1005.31
(1000)(170 )
= 0.006
415 2 m A = 0.90(0.006)(1000)(170) (415) 1− 0.59(0.006) 20
m A = 60007594.28N .mm m A = 60.01KN .m / m
Determine m* Cnit strip with bars perpendicular to the positive yield line between supports
19
mC = φρ bd 2 fy (1 − 0.59 ρ As =
1000 π 275 4
(16 )
As = 731.13mm
ρ =
fy fc′
)
2
2
731.13
(1000)(170 )
= 0.0043
mC = 0.90 ( 0.0043)(1000)( 170)
2
( 415) 1− 0.59( 0.0043)
415 20
mC = 4.3971439.91N .mm mC = 43.97 KN .m / m ∴
1 1 1 )(2.4 ) W i = (60.01)( )(2.4) + (43.97 )( + x x 3.6 − x 144.024 105.528 10.5.528 W i = + + (3.6 − x ) x x
W i =
249.55
x
+
105.528 3.6 − x
*ompute e&ternal wor#
20
1
W e = W u ( x )(2.4 )
+ W u (3.6 − x )(2.4 )
2 W e = 1.2W u X + 1.2W u (3.6 − x )
1 2
W e = 4.32W u W i = W e 277.296 117.264
x 249.55
+
= 4.32W u
3.6 − x 105.528
(3.6 − x )
x
4.32 57.766 24.428 +
x W u =
(3.6 − x )
57.766
+
x
= W u = W u
24.428
→ (1)
(3.6 − x )
To obtain the critical value of +u, find the value of & that minimizes +u. 0ind the first derivative of +u with respect to & then set to zero. dW u
=
x (0 ) − 57.766(1)
( x )
2
dx dW u
=
− 57.766
dx O=
x
2
− 57.766
+
(3.6 − x )(0 ) − 24.428(0 − 1) (3.6 − x )2
24.428
(3.6 − x )2
24.428
(3.6 − x )2 2 2 2 O = −57.766(3.6 − x ) + 24.428( x ) + 24.428 x x
2
+
+
O = −748.65 + 415.92 x − 57.766 x 2 + 24.428 x 2 O = −748.65 + 415.92 x − 33.338 x
2
2
33.338 x − 415.92 x + 748.65 = 0
x = x = x (+ )
− (− 415.92 )+
(415.92)2 − 4(33.338)(748.65) 2(33.338)
415.92+ 270.473 66.676 = 10.294, discard
x (− ) = 2.181
21
∴
x = 2.181m
ubstitute in equation 1 W U =
57.766 2.181
W U = 43.70
+
24.428
(3.6 − 2.181)
KN
M 2 W U = 43.70 KPa
3ote5 In this problem, it so happened that the resulting equation is easy to solve. 0or other situations, successive appro&imations may be done to obtain the solution. Problem 1
simply supported twoEway slab is reinforced in the short direction with in 12mmφ bars spaced F 1!) mm o.c., and 12mmφ bars F 1)4mm o.c. in the long direction. Determine the magnitude of the uniformly distributed failure load producing the yield line pattern shown. f c ′ = 20 MPa fy = 275 MPa
22
3?T$5 This given pattern may not be the critical pattern at failure. olution5 0ind the fle&ural strengths in the G and Y directions *onsider a unit strip of slab in both directions
23
H?T DI$*TI?3 6( m span; bars parallel to the yEa&is7
As =
ρ =
1000 π
(12)2
125 4 904.78
(1000)(124)
= 904.78mm
2
= 0.007
3?T$5
fy
fc′
m X = φρbd S 2 fy 1 − 0.59ρ m x = ( 0.007 )(1000)(124)
2
( 275) ( 275) 1− 0.59( 0.007)
20
m X = 25126162.34N .mm m X = 25.13KN .m / m
"?3 DI$*TI?3 6 m span; bars parallel to the &Ea&is7
24
As =
ρ =
1000 π
(12)2
150 4 753.98
(1000)(112)
= 753.98mm
2
= 0.0067
3?T$5
m y = 0.90 ( 0.0067)(1000)( 112)
2
( 275) ( 275) 1− 0.59( 0.0067)
20
m y = 19670470.87 N .mm m y = 19.67 KN .m / m
*onsidering the inclined yield line, the moment normal to the line is5 mα = m X cos 2 α + my sin 2 α
mα = 25.13cos
2
( 45 ) + 19.67sin ( 45 ) 0
2
0
mα = 22.40 KN .m
Determine yield line rotations
25
s before, assume a virtual deflection of ∆ = 1 at the center of the slab at point f and along as shown. ?bviously, in this case, a : b : d. Aoreover, note that d is drawn simply as an e&tension of a in order to determine the rotation.
otation of inclined yield line b 6due to symmetry, the rotation is the same for all inclined yield lines7
+here5 a = d = 2 2 1 1 θ af = = a 2 2
θ df =
1
d
=
1 2 2
θ ad : total rotation of inclined yield line
26
θ ad = θ ad =
1 2 2 1+1
θ ad = θ ad = θ ad = θ ad
1
+
2 2 2
=
2
2 2
1 2 1
⋅
2
2 2
2
2 = 0.707radian
ince all inclined lines have the same rotations, let; θ i = 0.707 radian for each inclined line otation of horizontal yield line, θ h
θ h = θ L + θ R θ h =
1
2 θ h = 1
+
1 2
*ompute internal wor# done Wi = ∑ mlθ Wi = mi l iθ i + m X l hθ h
27
+here5 l i = Total length of inclined yield lines l h = "ength of horizontal yield lines
mi = mα li =
8
lh =
2m
W i = 4 ( 22.40 )
( 8 ) ( 0.707) + ( 25.13)( 2)( 1)
Wi = 229.374KN .m
0ind e&ternal wor# done "et5 +u: magnitude uniform load over the slab in each segment. The resultant of +u, passes through the centroid of the corresponding segment, as shown in the f ollowing figure identified as points 1,!,',(,),,- and /. The corresponding deviations8deflections of those points are >1, >!, >', >(, >), >, >-, and >/, with resultants 1, !, ', (, ), , -, and /, respectively.
esultant = (W U ) 6area7
28
1 (2)(2) = 2W U 2
R1 = W U ∆1 =
1
∆=
3
1 3
(1) =
1 3
1 (4)(2) = 4W U 2
R2 = W U ∆2 =
1
∆=
3
1 3
1 (2)(2) = 4W U 2
R3 = W U ∆3 =
1
3 R4 = W U [(2)(2)] = 4W U 1
(1) = 0.5 2 R5 = 4W U ∆4 =
∆ 5 = 0.5
R6 = 2W U ∆6 =
1
3 R7 = 4W U ∆7 =
1
3 R8 = 2W U ∆8 =
1
3 W e = R1 ∆1 + R2 ∆ 2 + R3 ∆ 3 + R4 ∆ 4 + R5 ∆ 5 + R6 ∆ 6 + R7 ∆ 7 + R8 ∆ 8
1 1 1 1 W e = 2W U + 4W U + 2W U + 4W U (0.5) + 4W U (0.5) + 2W U 3 3 3 3 1 1 + 2W U 3 3
+ 4W U
W e = 9.33W U
29
