Yield Line Method

YIELD LINE METHOD For an under-reinforced section the capacity to develop curvatures between the first yield of reinforcement and failure due to crushing of concrete is considerable. For a slab which is subjected to increasing load, cracking and reinforcement yield will first occur in the most highly stressed zone. This will then act as a plastic hinge as subsequent loads are distributed to other regions of the slab. Cracks will develop to form a pattern of 'yield lines' until a mechanism is formed and collapse is indicated by increasing deflections under constant load.

It is assumed that a pattern of yield lines can be superimposed on the slab, which will cause a collapse mechanism, and that the regions between yield lines remain rigid and un-cracked. Figure 1 show the yield line mechanism which will occur for the simple case of a fixed ended slab spanning in one direction with a uniform load. Rotation along the yield lines will occur at a constant moment equal to the ultimate moment of resistance of the section, and will absorb energy. This can be equated to the energy expended by the applied load undergoing a compatible displacement and is known as the virtual work method

Yield line

Fixed supprt

Plastic hinges

Considerable care must be taken over the selection of likely yield line patterns since the method will give an ‘upper bound’ solution, that is, either a correct or unsafe solution. Yield lines will form at right angles to bending moments which have reached the ultimate moment of resistance of the slab and the following rules may be helpful: (a) Yield lines are usually straight and end at a slab boundary. (b) Yield lines will lie along axes of rotation, or pass through their points of intersection. (c) Axes of rotation lie along supported edges pass over columns or cut unsupported edges.

Few patterns of yield lines:

Figure 2 A yield line caused by a sagging moment is generally referred to as a ‘positive’ yield line and represented by a full line. While a hogging moment causing cracking on the top of the slab causes a ‘negative’ yield line shown by a broken yield line.

Virtual Work Method Once the yield line have been chosen, some point on the slab is given a virtual displacement, δ as shown in Figure 3. Δc

θ

δ 2θ

Figure 3

The external work done by the load when displaced this amount is External work =    w dx dy where

= Σ (WΔc) w – load on an element of area δ – deflection of that element W – total load on a plate segment Δc – deflection of the centroid of that segment

The total external work is the sum of the work for each plate. The internal work done by rotating the yield line is Internal work = Σ(mblθ) where

mb – bending moment per unit length of yield line l – length of the yield line θ – angle change at that yield line

The total internal work done during the virtual displacement is the sum of the internal work done on each yield line. The principle of virtual work states that for equilibrium External work = internal work Σ (WΔc) = Σ(mblθ)

Yield line analysis for one way slab

a) Cross section C

A

E

b

B

D

F

b) Plan

θ

δ

Δc mx1

mx1 mx2

2θ mx2

c) Deformed shape

Figure 4 The one way slab shown in the Figure 4 has reinforcement at the top at the ends with a moment capacity mx1 per unit width and reinforcement at the bottom of the mid-span with capacity mx2 per unit width. It is loaded with a uniform load of w per unit area. Compute the moment capacity required to support w. Step 1- select axes and yield lines. Axes and negative moment yield lines will form along the faces of the supports and a positive moment yield line will form at a mid-span.

Step 2 – Give a slab a virtual displacement Some point in the slab is displaced downward by an amount δ. Line C-D had been displaced δ. Step 3 – Compute the external work The total load on plate segment A-B-C-D is W = w(bL/2). The displacement of the resultant load W : Δc = δ/2. Therefore the external work done on plate A-B-C-D is W Δc = w

bL    wbL   2 2 4

The total external work for the two plate segment is wbL  External work = 2  

4



Step 4 – Compute the internal work The negative moment yield line at A-B rotates through an angle θ, where θ = δ/ (L/2). The external work done in rotating through this angle is mx1b(2δ/L) The total internal work done at all three yield lines is 

2 

4

Internal work = 2m x1b   m x 2 b   L   L   Step 5 – Equate the external and internal work   2  wbL  2  = 2m x1b  L  4  

  4    m x 2 b    L 

Thus m x1  m x 2 

wL2 8

Any combination of mx1 and mx2 that equal wL2/8 will satisfy equilibrium.

Square slab The simply supported square slab has a positive moment capacity, mx = my. Compute the value of m required to support a uniform load w. Steps 1 – select the axes and yield lines Yield lines will form along the diagonal which as shown in Figure 5 are lines of the maximum moment. The segments will rotate about axes along the four supports. L A

L

1

G

D

E

F

B

C

1

L

L

2

2

B θ1

θ2

δ E

D

θ

Section 1-1

Figure 5 Step 2 – give a slab a virtual displacement Point E is displaced downward by an amount of δ. Step 3 – Compute the external work Load on plate segment A-B-E

L2 W= w 4

Deflection of centroid of plate

c 



3

External work done on plate A-B-E Total external work

 wL2    W c    4 3  wL2     W c  4 4  3

Step 4 – Compute the internal work Consider the yield line A-E. Length l = L/√2. The rotation of yield line A-E is   1   2

Where 1  2



2 2

L

 2

 L

The internal work on line A-E = mblθ and mb is equal to mx and my. Thus mblθ = m(L/√2)(2√2δ/L) = 2mδ. The total work on all four yield lines is (mb l )  4(2m )

Step 5 – equate the external and internal work wL2 wL2  8m and m  24 3

Thus the reinforcement in both direction should be designed for L2 mrx  mry  w f 24

Where wf is the factored uniform load

Calculation of the load capacity of a reinforced concrete slab 1.5 m

1.5 m

1.5 m

m1= 10 kNm/m m2 = 20 kNm/m

m3 = 15 kNm/m m1

m1 3m

m3 m1

m2

(a) Reinforcement at top of slab m4= 5 kNm/m

0.5 m

m5= 5 kNm/m

2m

0.5 m

3.5 m

0.5 m

0.5 m

(b) Reinforcement at bottom of slab

A

B

F

E

D

C (c) Assumed yield line pattern

Figure 6

Figure 6 shows a rectangular slab that is fixed on four sides and has negative moment (top) reinforcement with the capacities shown in Figure 6(a), and positive moment (bottom) reinforcement as shown in Figure 6(b). Compute the load w corresponding to failure of this slab. Step 1 – Select axes and yield lines Axes will form along the four sides of the slab. Step 2 – Give a slab a virtual displacement Line E – F is given a virtual displacement of δ Step 3 – Compute the external work Panel A-C-E  1.5   W c  w 3x  = 0.75wδ kNm 2 3 

Panel A-B-F-E   1.5   W C  w(1.5  1.5)   ( w  1.5  1.5)  1.875w kNm 2  2 3

Total external work = 2 x 0.75wδ + 2 x 1.875 wδ = 5.25 wδ kNm Step 4 – Compute the internal work Panel A-C-E mxLyθy = (15 x 3 x δ/1.5) + (5 x 2 x δ/1.5) = 36.67δ kNm Panel A-B-F-E mYLXθX = (10 x (1.5 + 1.5) x δ/1.5) + (20 x 1.5 x δ/1.5) + (5 x 3.5 x δ/1.5) = 51.678δ kNm Total internal work = 2 x 36.678 + 2 x 51.678 = 176.78δ kNm Step 5 – equate the external and internal work 5.25 wδ = 176.78 δ

so w = 33.6 kN/m2