010 chapter 4 - INFLIBNETshodhganga.inflibnet.ac.in/bitstream/10603/12388/10/10_chapter 4.pdf · Various support conditions of slab (Nine edge conditions). In addition to the above,

80

CHAPTER 4

APPLICATION OF WORK EQUATIONS AND THEIR

PARAMETRIC STUDY

4.1. Introduction:

In the last chapter the work equations are derived for corner openings of

orthogonal slabs and can be applied theoretically to various problems and a

parametric study on strength of orthogonal slab with openings will be made with the

following data:

1. The aspect ratio of slab and opening (r, and respectively).

2. Location of opening

3. The Coefficient of Orthotrophy ().

4. Coefficients of Orthogonal Moments ( 'xK , '

yK , I1, I2, I3 and I4).

5. Sum of the Coefficients of Orthogonal Moments (K).

6. Various support conditions of slab (Nine edge conditions).

In addition to the above, a comparative study can be made in strengths of

orthogonal slab with openings with solid slab. The valid admissible governing failure

pattern can also be found.

4.2 Importance of choosing different moment coefficients:

To obtain m

wLy2

values the work equations derived in the previous chapter are

to be evaluated for different parameters given in the section 4.1. The work so for

81

published by different researchers2,8,11,14,17

predominantly concludes that the value of

m

wLy2

mainly depends on the selection of orthogonal moment coefficients ( 'xK ,

'

yK ,

I1, I2, I3 and I4), coefficient of orthotropy () and sum of orthogonal moment

coefficients (K) for a given specific problem. Hence the author has chosen three

different varieties of orthogonal moment coefficients, one basing on affined theorm3,14

and other two basing on individual designers choice as given in appendix 1.

4.3 Segment Equilibrium Method:

To verify the value obtained by using virtual work method, the segment

equilibrium method has been used for various problems of slabs with and without

opening for various parameters mentioned in section 4.1. To solve a slab by

equilibrium method, the steps suggested by Park, R. and W.L. Gamble14

are followed.

The section below gives a comparative study of virtual work and segment equilibrium

method for a solid slab.

4.4 Shifting of failure patterns in solid slab:

Problem 4.1: A continuous solid slab whose aspect ratio is 1.2, is provided

with following orthogonal moment coefficients as given below. (Refer article 4.6)

'

xK = 0.33, '

yK = 1.0, I1(I3) = 1.0, I2(I4) = 1.67, K = 4.0, = 0.5 (designers choice

coefficients)

Problem 4.2: A two adjacent edges continuous solid slab whose aspect ratio is

1.2, is provided with following orthogonal moment coefficients as given below.

(Refer article 4.7)

82

'

xK = 0.9, '

yK = 0.5, I1(I3) = 1.5, I2(I4) = 1.1, K = 4.0, = 1.5 (designers

choice coefficients)

The solutions obtained by computer programme and segmental equilibrium

method are shown in the following pages.

The solution of these two problems clearly indicate that the choice of selecting

orthogonal moment coefficients is very much essential in order to have the desired

strength and failure pattern. Shifting of failure patterns in solid slab for other edge

conditions (O.S.D, O.L.D, O.S.C, O.L.C, T.S.C, T.L.C, S.S) are explained in the

Article Nos. 4.8, 4.9, 4.10, 4.11, 4.12, 4.13, 4.14. The same idea has been extended

to slabs with corner openings.

4.5 Shifting of failure patterns in slab with corner opening:

Problem 4.3: A continuous solid slab whose aspect ratio is 1.0, is provided

with three different sets of orthogonal moment coefficients as given below. (Refer

Article Nos. 4.15, 4.16 and 4.17)

Set 1: 'x

K = 0.5, '

yK = 1.0, I1(I3) = 0.5, I2(I4) = 1.0, K = 3.0, = 0.5 (affine theorem

coefficients)

Set 2: 'x

K = 0.3, '


choice coefficients)

Set 3: 'x

K = 0.7, '


choice coefficients reversed)

83

Problem 4.4: A two adjacent edges continuous solid slab whose aspect ratio is

1.0, is provided with following orthogonal moment coefficients as given below.

(Refer Article Nos. 4.18, 4.19 and 4.20)

Set 1: 'x

K = 0.8, 'y

K = 1.2, I1(I3) = 0.8, I2(I4) = 1.2, K = 4.0, = 0.667 (affine theorem

coefficients)

Set 2: 'x

K = 0.6, '

yK = 1, I1(I3) = 1, I2(I4) = 1.4, K = 4.0, = 1.0 (designers choice

coefficients)

Set 3: 'x

K = 1.0, '


choice coefficients reversed)

The solutions obtained by computer programme and segmental equilibrium

method are shown in the following pages.

The solution of these two problems clearly indicate that the choice of selecting

orthogonal moment coefficients is very much essential in order to have the desired

strength and failure pattern. Shifting of failure patterns in slab with corner opening

for other edge conditions of the slab (O.S.D, O.L.D, O.S.C, O.L.C, T.S.C, T.L.C and

S.S) are explained in the Article Nos. 4.21, 4.22, 4.23, 4.24, 4.25, 4.26, 4.27, 4.28,

4.29, 4.30, 4.31, 4.32, 4.33, 4.34, 4.35, 4.36, 4.37, 4.38, 4.39 and 4.40 respectively.

84

4.6. CONTINUOUS SLAB (C.S.) (Refer Fig. 4.1)

r = 1.2

= 0.5, k = 4.0

Virtual work method: Designers choice coefficients

m

wLy2

= 41.60588

r1 = 2.73986 c1 = 4.4

r2 = 2.73986 c2 = 4.4

r3 = 2.00 c3 = 5.0

Failure Pattern: I

Segmental equilibrium method:

Segment A:

0.5 10 4.42 w/3

= 10 0.33m + 10 1.0m m

wLy2

= 41.21

Segment B:

0.5 4.4 52 w/3 + 3.2 52 w/2 + 0.5 4.4 52 w/3

= 12 1.0m + 12 1.67m m

wLy2

= 41.78

Segment C:

0.5 10 4.42 w/3

= 10 0.33m + 10 1.0m m

wLy2

= 41.21

Segment D:

0.5 4.4 52 w/3 + 3.2 52 w/2 + 0.5 4.4 52 w/3

= 12 1.0m + 12 1.67m m

wLy2

= 41.78

Average m

wLy2

= 41.50

1.0

m

0.3

3 m

1.0 m

1.67 m

86

4.7. TWO ADJACENT EDGES CONTINUOUS SLAB (T.A.C.) (Refer Fig. 4.2)

r = 1.2

= 1.5, k = 4.0


m

wLy2

= 24.65619

r1 = 1.61237 c1 = 7.44

r3 = 1.59987 c3 = 6.3

r4 = 2.86704 c4 = 3.5

Failure Pattern: 2


Segment A:

0.5 3.5 7.442 w/3 + 0.20 7.442 w/2 + 0.5 6.3 7.442 w/3

= 10 0.9m + 10 1.5m m

wLy2

= 25.0140

Segment B:

0.5 12 6.32 w/3

= 12 0.5m + 12 1.1m m

wLy2

= 24.1875

Segment C:

0.5 6.3 4.562 w/3 + 0.20 4.562 w/2 + 0.5 3.5 4.562 w/3

= 10 0.9m m

wLy2

= 24.9707

Segment D:

0.5 12 3.52 w/3

= 12 0.5m m

wLy2

= 24.4898

Average m

wLy2

= 24.6655

1.5

m

0.9

m

0.5 m

1.1 m

88

4.8. ONE SHORT EDGE DISCONTINUOUS SLAB (O.S.D.) (Refer Fig. 4.3)

r = 1.2

= 0.667, k = 4.0


m

wLy2

= 35.80807

r1 = 2.31844 c1 = 5.2

r2 = 3.78986 c2 = 3.2

r3 = 2.00 c3 = 5.0

Failure Pattern: I


Segment A:

0.5 10 5.22 w/3

= 10 0.6m + 10 1.0m m

wLy2

= 35.50

Segment B:

0.5 5.2 52 w/3 + 3.6 52 w/2 + 0.5 3.2 52 w/3

= 12 1.0m + 12 1.4m m

wLy2

= 36.0

Segment C:

0.5 10 3.22 w/3

= 10 0.6m m

wLy2

= 35.15

Segment D:

0.5 5.2 52 w/3 + 3.6 52 w/2 + 0.5 3.2 52 w/3

= 12 1.0m + 12 1.4m m

wLy2

= 36.0

Average m

wLy2

= 35.6625

1.0

m

0.6

m

1.0 m

1.4 m

90

4.9. ONE LONG EDGE DISCONTINUOUS SLAB (O.L.D) (Refer Fig. 4.4)

r = 1.2

= 1.0, k = 4.0


m

wLy2

= 31.01776

r1 = 2.0 c1 = 6.0

r3 = 1.60987 c3 = 6.2

r4 = 2.9397 c4 = 3.4

Failure Pattern: 2


Segment A:

0.5 3.4 62 w/3 + 0.4 62 w/2 + 0.5 6.2 62 w/3

= 10 0.5m + 10 1.5m m

wLy2

= 30.8642

Segment B:

0.5 12 6.22 w/3

= 12 0.6m + 12 1.4m m

wLy2

= 31.2175

Segment C:

0.5 6.2 62 w/3 + 0.4 62 w/2 + 0.5 3.4 62 w/3

= 10 0.5m + 10 1.5m m

wLy2

= 30.8642

Segment D:

0.5 12 3.42 w/3

= 12 0.6m m

wLy2

= 31.1419

Average m

wLy2

= 31.0219

1.5

m

0.5

m

0.6 m

1.4 m

92

4.10. ONE SHORT EDGE CONTINUOUS SLAB (O.S.C) (Refer Fig. 4.5)

r = 1.2

= 0.667, k = 4.0


m

wLy2

= 35.09861

r1 = 2.29259 c1 = 5.2

r2 = 4.09986 c2 = 2.9

r3 = 2.00 c3 = 5

Failure Pattern : 1


Segment A:

0.5 10 5.22 w/3

= 10 0.5m + 10 1.1m m

wLy2

= 35.5029

Segment B:

0.5 5.2 52 w/3 + 3.9 52 w/2 + 0.5 2.9 52 w/3

= 12 2.4m m

wLy2

= 34.9091

Segment C:

0.5 10 2.92 w/3

= 10 0.5m m

wLy2

= 35.6717

Segment D:

0.5 5.2 52 w/3 + 3.9 52 w/2 + 0.5 2.9 52 w/3

= 12 2.4m m

wLy2

= 34.9091

Average m

wLy2

= 35.2482

1.1

m

0.5

m

2.4 m

94

4.11. ONE LONG EDGE CONTINUOUS SLAB (O.L.C.) (Refer Fig. 4.6)

r = 1.2

= 0.667, k = 4.0


m

wLy2

= 29.49768

r1 = 2.09920 c1 = 5.7

r2 = 2.09920 c2 = 5.7

r3 = 1.5 c3 = 6.67

Failure Pattern : 1


Segment A:

0.5 10 5.72 w/3

= 10 1.6m m

wLy2

= 29.5476

Segment B:

0.5 5.7 6.672 w/3 + 0.60 6.672 w/2 + 0.5 5.7 6.672 w/3

= 12 0.6m + 12 1.8m m

wLy2

= 29.4251

Segment C:

0.5 10 5.72 w/3

= 10 1.6m m

wLy2

= 29.5476

Segment D:

0.5 5.7 3.332 w/3 + 0.60 3.332 w/2 + 0.5 5.7 3.332 w/3

= 12 0.6m m

wLy2

= 29.5135

Average m

wLy2

= 29.5085

1.6

m

0.6 m

1.8 m

96

4.12. TWO SHORT EDGES CONTINUOUS SLAB (T.S.C) (Refer Fig. 4.7)

r = 1.2

= 0.5, k = 4.0


m

wLy2

= 41.70639

r1 = 2.72808 c1 = 4.4

r2 = 2.72808 c2 = 4.4

r3 = 2.0 c3 = 5.0

Failure Pattern : 1


Segment A:

0.5 10 4.42 w/3

= 10 0.54m + 10 0.8m m

wLy2

= 41.5246

Segment B:

0.5 4.40 52 w/3 + 3.20 52 w/2 + 0.5 4.40 52 w/3

= 12 2.67m m

wLy2

= 41.7895

Segment C:

0.5 10 4.402 w/3

= 10 0.54m + 10 0.8m m

wLy2

= 41.5246

Segment D:

0.5 4.40 52 w/3 + 3.20 52 w/2 + 0.5 4.40 52 w/3

= 12 2.67m m

wLy2

= 41.7895

Average m

wLy2

= 41.6571

0.8

m

0.5

4 m

2.67 m

98

4.13. TWO LONG EDGES CONTINUOUS SLAB (T.L.C) (Refer Fig. 4.8)

r = 1.2

= 1.5, k = 4.0


m

wLy2

= 39.19795

r1 = 2.0 c1 = 6

r3 = 2.01986 c3 = 4.9

r4 = 2.01986 c4 = 4.9

Failure Pattern : 2


Segment A:

0.5 4.90 62 w/3 + 0.2 62 w/2 + 0.5 4.90 62 w/3

= 10 2.4m m

wLy2

= 38.4615

Segment B:

0.5 12 4.902 w/3

= 12 0.5m + 12 1.1m m

wLy2

= 39.9833

Segment C:

0.5 4.90 62 w/3 + 0.2 62 w/2 + 0.5 4.90 62 w/3

= 10 2.4m m

wLy2

= 38.4615

Segment D:

0.5 12 4.902 w/3

= 12 0.5m + 12 1.1m m

wLy2

= 39.9833

Average m

wLy2

= 39.2224

2.4

m

0.5 m

1.1 m

100

4.14. SIMPLY SUPPORTED SLAB (S.S) (Refer Fig. 4.9)

r = 1.2

= 0.5, k = 4.0


m

wLy2

= 41.60588

r1 = 2.73986 c1 = 4.4

r2 = 2.73986 c2 = 4.4

r3 = 2.0 c3 = 5

Failure Pattern : 1


Segment A:

0.5 10 4.42 w/3

= 10 1.33m m

wLy2

= 41.2148

Segment B:

0.5 4.4 52 w/3 + 3.2 52 w/2 + 0.5 4.4 52 w/3

= 12 2.67m m

wLy2

= 41.7895

Segment C:

0.5 10 4.42 w/3

= 10 1.33m m

wLy2

= 41.2148

Segment D:

0.5 4.4 52 w/3 + 3.2 52 w/2 + 0.5 4.4 52 w/3

= 12 2.67m m

wLy2

= 41.7895

Average m

wLy2

= 41.5022

1.3

3 m

2.67 m

102

4.15. CONTINUOUS SLAB (C.S) (Refer Fig. 4.10)

= 0.3, = 0.2 r = 1.0

= 0.5, k = 3.0

Virtual work method: Affine theorem coefficients

m

wLy2

= 32.28133

r1 = 2.3 c1 = 4.4

r2 = 2.36923 c2 = 4.2

r3 = 2.2 c3 = 4.5

Failure Pattern: 2


Q1 = 2.44/2.5 1.0m = 0.976m

Segment A:

2.5 1.962 w/2 + 0.5 2.5 2.44 (1.96 + 2.44/3)w

+ 0.5 5.5 4.42 w/3 0.976m 1.96 = 8 0.5m + 8 0.5m

m

wLy2

= 31.9699

Segment B:

0.5 2.44 2.5 (2+2.5/3) w + (4.4 3) 22 w/2 + 1.4 4.52 w/2

+ 0.5 4.2 4.52 w/3 + 0.976m 2 = 7 1.0m + 8.04 1.0m

m

wLy2

= 32.8914

Segment C:

0.5 10 4.22 w/3 = (10 0.5m) + (10 0.5m)

m

wLy2

= 34.01

Segment D:

0.5 4.4 5.52 w/3 + 1.4 5.52 w/2 + 0.5 4.2 5.52 w/3

= 10 1.0m + 10 1.0m m

wLy2

= 30.99

Average m

wLy2

= 32.4653

0.5

m

0.5

m

1.0 m

1.0 m

104


= 0.3, = 0.2

r = 1.0

= 0.5, k = 3.0


m

wLy2

= 32.14192

r1 = 2.4 c1 = 4.2

r2 = 2.31429 c2 = 4.3

r3 = 2.2 c3 = 4.5

Failure Pattern: 2


Q1 = 2.33/2.5 0.8m = 0.746m

Segment A:

2.5 1.872 w/2 + 0.5 2.5 2.33 (1.87 + 2.33/3)w

+ 0.5 5.5 4.22 w/3 0.746m 1.87 = 8 0.3m + 8 0.7m

m

wLy2

= 33.2573

Segment B:

0.5 2.33 2.5 (2+2.5/3) w + 1.2 22 w/2 + 1.5 4.52 w/2

+ 0.5 4.3 4.52 w/3 + 0.746m 2.0 = 7 1.2m + 8.13 0.8m

m

wLy2

= 33.2375

Segment C:

0.5 10 4.32 w/3 = (10 0.3m) + (10 0.7m)

m

wLy2

= 32.4465

Segment D:

0.5 4.2 5.52 w/3 + 1.5 5.52 w/2 + 0.5 4.3 5.52 w/3

= (10 0.8m) + (10 1.2m) m

wLy2

= 30.515

Average m

wLy2

= 32.3641

0.7

m

0.3

m

0.8 m

1.2 m

106


= 0.3, = 0.2

r = 1.0

= 0.5, k = 3.0

Virtual work method: Reverse of Designers choice coefficients

m

wLy2

= 32.41597

r1 = 2.2 c1 = 4.5

r2 = 2.3333 c2 = 4.3

r3 = 2.2 c3 = 4.5

Failure Pattern: 2

Segmental Equilibrium method:

Q1 = 2.5/2.5 1.2m = 1.2m

Segment A:

22 2.5 w/2 + 0.5 2.5 2.5 (2 + 2.5/3)w

+ 0.5 5.5 4.52 w/3 1.2m 2.0 = 8 0.3m + 8 0.7m

m

wLy2

= 32.0823

Segment B:

0.5 2.5 2.5 (2 + 2.5/3) w + 1.5 22 w/2 + 1.2 4.52 w/2

+ 0.5 4.3 4.52 w/3 + 1.2m 2 = 7 0.8m + 8 1.2m

m

wLy2

= 33.2324

Segment C:

0.5 10 4.32 w/3 = (10 0.7m) + (10 0.3m)

m

wLy2

= 32.4499

Segment D:

0.5 4.5 5.52 w/3 + 1.2 5.52 w/2 + 0.5 4.3 5.52 w/3

= 10 1.2m + 10 0.8m m

wLy2

= 31.9915

Average m

wLy2

= 32.4390

0.3

m

0.7

m

1.2 m

0.8 m

108

4.18. TWO ADJACENT EDGES CONTINUOUS SLAB (T.A.C) (Refer Fig. 4.13)

= 0.1, = 0.3

r = 1.0

= 0.667, k = 4.0


m

wLy2

= 31.97596

r1 = 2.13 c1 = 4.7

r2 = 2.58495 c2 = 3.9

r3 = 1.71 c3 = 5.8

Failure Pattern: 3


Q1 = 4.566/3.7 0.8m = 0.987m

Segment A:

0.5 4.566 3.7 (1 + 3.7/3) w + 2.8 12 w/2

+ 0.5 4.2 4.72 w/3 + 0.987m 1 = 7 0.8m + 8.766 0.8m

m

wLy2

= 32.5396

Segment B:

0.5 3.7 4.566 (1.234 + 4.566/3) w + 3.7 1.2342 w/2 + 1.4 5.82

w/2

+ 0.5 5.82 3.9 w/3 0.987m 1.234 = 9 1.2m + 9 1.2m

m

wLy2

= 31.9081

Segment C:

0.5 10 3.92 w/3 = (10 0.8m)

m

wLy2

= 31.5581

Segment D:

0.5 4.7 4.22 w/3 + 1.4 4.22 w/2 + 0.5 3.9 4.22 w/3

= (10 1.2m) m

wLy2

= 31.8877

Average m

wLy2

= 31.9733

0.8

m

0.8

m

1.2 m

1.2 m

110


= 0.1, = 0.3

r = 1.0

= 0.667, k = 4.0


m

wLy2

= 28.88849

r1 = 2.03 c1 = 4.9

r2 = 2.87087 c2 = 3.5

r3 = 1.71 c3 = 5.8

Failure Pattern: 3


Q1 = 4.616/3.9 0.6m = 0.71m

Segment A:

2.8 12 w/2 + 0.5 4.616 3.9 (1 + 3.9/3)w

+ 0.5 4.2 4.92 w/3 + 0.71m 1 = 7 1.0m + 8.816 0.6m

m

wLy2

= 29.7601

Segment B:

0.5 4.616 3.9 (1.184 + 4.616/3) w + 3.9 1.842 w/2 + 1.6 5.82

w/2

+ 0.5 3.5 5.82 w/3 0.71m 1.184 = 9 1.4m + 9 1.0m

m

wLy2

= 28.9018

Segment C:

0.5 10 3.52 w/3 = (10 0.6m)

m

wLy2

= 29.3877

Segment D:

0.5 4.9 4.22 w/3 + 1.6 4.22 w/2 + 0.5 3.5 4.22 w/3

= (10 1.0m) m

wLy2

= 25.7678

Average m

wLy2

= 28.4544

1.0

m

0.6

m

1.0 m

1.4 m

112


= 0.1, = 0.3

r = 1.0

= 0.667, k = 4.0


m

wLy2

= 34.93547

r1 = 2.23 c1 = 4.5

r2 = 2.41301 c2 = 4.1

r3 = 1.810 c3 = 5.5

Failure Pattern: 3


Q1 = 4.278/3.5 1m = 1.222m

Segment A:

0.5 4.278 3.5 (1 + 3.5/3) w + 2.5 12 w/2

+ 0.5 4.5 4.52 w/3 + 1.222m 1 = = 7 0.6m + 8.778 1.0m

m

wLy2

= 35.997

Segment B:

0.5 3.5 4.278 (1.222 + 4.278/3) w + 3.5 1.2222 w/2 + 1.4 5.52

w/2

+ 0.5 4.1 5.52 w/3 1.222m 1.222 = 9 1.0m + 9 1.4m

m

wLy2

= 35.9242

Segment C:

0.5 10 4.12 w/3 = (10 1.0m)

m

wLy2

= 35.693

Segment D:

0.5 4.5 4.52 w/3 + 1.4 4.52 w/2 + 0.5 4.1 4.52 w/3

= (10 1.4m) m

wLy2

= 32.4074

Average m

wLy2

= 35.0054

0.6

m

1.0

m

1.4 m

1.0 m

114

4.21. ONE SHORT EDGE DISCONTINUOUS SLAB (O.S.D) (Refer Fig. 4.16)

= 0.3, = 0.2

r = 1.0

= 0.5, k = 4.0


m

wLy2

= 38.02753

r1 = 2.1 c1 = 4.8

r2 = 3.10909 c2 = 3.2

r3 = 2.2 c3 = 4.5

Patterns: 2


Q1 = 2.67/2.5 1.33m = 1.4204m

Segment A:

0.5 2.5 2.67 (2.13 + 2.67/3)w + 2.5 2.132 w/2

+ 0.5 5.5 4.82 w/3 1.4204m 2.13

= 8 0.67m + 8 0.67m m

wLy2

= 37.2804

Segment B:

0.5 2.67 2.5 (2 + 2.5/3) w + 1.8 22 w/2 + 2 4.52 w/2

+ 0.5 3.2 4.52 w/3 + 1.4204m 2

= 7 1.33m + 7.87 1.33m m

wLy2

= 38.3989

Segment C:

0.5 10 3.22 w/3 = (10 0.67m)

m

wLy2

= 39.2577

Segment D:

0.5 4.8 5.52 w/3 + 2 5.52 w/2 + 0.5 3.2 5.52 w/3

= (10 1.33m) + (10 1.33m) m

wLy2

= 37.6877

Average m

wLy2

= 38.1562

0.6

7 m

0.6

7 m

1.33 m

1.33 m

116


= 0.3, = 0.2

r = 1.0

= 0.5, k = 4.0


m

wLy2

= 34.58847

r1 = 2.1 c1 = 4.8

r2 = 4.20909 c2 = 2.4

r3 = 2.2 c3 = 4.5

Patterns: 2


Q1 = 2.67/2.5 1.0m = 1.068m

Segment A:

0.5 2.5 2.67 (2.13 + 2.67/3)w + 2.5 2.132 w/2

+ 0.5 5.5 4.82 w/3 1.068m 2.13

= 8 0.33m + 8 1.0m m

wLy2

= 35.0278

Segment B:

0.5 2.67 2.5 (2 + 2.5/3) w + 1.8 22 w/2 + 2.8 4.52 w/2

+ 0.5 2.4 4.52 w/3 + 1.068m 2

= 7 1.67m + 7.87 1.0m m

wLy2

= 35.1956

Segment C:

0.5 10 2.42 w/3 = (10 0.33m)

m

wLy2

= 34.3750

Segment D:

0.5 4.8 5.52 w/3 + 2.8 5.52 w/2 + 0.5 2.4 5.52 w/3

= (10 1.0m) + (10 1.67m) m

wLy2

= 33.9479

Average m

wLy2

= 34.6366

1.0

m

0.3

3 m

1.0 m

1.67 m

118


= 0.3, = 0.2

r = 1.0

= 0.5, k = 4.0


m

wLy2

= 40.91521

r1 = 2.2 c1 = 4.5

r2 = 2.6333 c2 = 3.8

r3 = 2.2 c3 = 4.5

Failure Pattern: 2


Q1 = 2.5/2.5 1.67m = 1.67m

Segment A:

0.5 2.5 2.5 (2 + 2.5/3) w + 2.5 22 w/2

+ 0.5 5.5 4.52 w/3 1.67m 2 = 8 1.0m + 8 0.33m

m

wLy2

= 40.1261

Segment B:

0.5 2.5 2.5 (2+2.5/3) w + 1.5 22 w/2 + 1.7 4.52 w/3

+ 0.5 3.8 4.52 w/3 + 1.67m 2 = 7 1.0m + 8 1.67m

m

wLy2

= 43.0763

Segment C:

0.5 10 3.82 w/3 = (10 1.0m)

m

wLy2

= 41.5455

Segment D:

0.5 4.5 5.52 w/3 + 1.7 5.52 w/2 + 0.5 3.8 5.52 w/3

= (10 1.67m) + (10 1.0m) m

wLy2

= 39.5214

Average m

wLy2

= 41.0673

0.3

3 m

1.0

m

1.67 m

1.0 m

120


= 0.4, = 0.3

r = 1.0

= 0.667, k = 3.0


m

wLy2

= 26.75861

r1 = 2.0111 c1 = 5.0

r3 = 1.98147 c3 = 5.0

r4 = 2.21888 c4 = 4.5

Failure Pattern: 9


Q1 = 2/2 0.9m = 0.9m

Segment A:

2 32 w/2 + 0.5 2 2 (3 + 2/3) w + 0.5 52 w/2

+ 0.5 4.5 52 w/3 0.9m 3 = 7 0.6m + 7 0.6m

m

wLy2

= 26.8549

Segment B:

0.5 2 2 (3 + 2/3) w + 1 32 w/2 + 0.5 5 52 w/3

+ 0.9m 3 = 6 0.9m + 7 0.9m

m

wLy2

= 27.5507

Segment C:

0.5 5 52 w/3 + 0.5 52 w/2 + 0.5 4.5 52 w/3

= 10 0.6m + 10 0.6m m

wLy2

= 26.1837

Segment D:

0.5 10 4.52 w/3

= 10 0.9m m

wLy2

= 26.6667

Average m

wLy2

= 26.8140

0.6

m

0.6

m

0.9 m

0.9 m

122


= 0.4, = 0.3

r = 1.0

= 0.667, k = 3.0


m

wLy2

= 24.27114

r1 = 2.1111 c1 = 4.7

r3 = 1.8148 c3 = 5.5

r4 = 2.62729 c4 = 3.8

Failure Pattern: 9


Q1 = 2.136/2.5 0.6m = 0.513m

Segment A:

2.5 2.5642 w/2 + 0.5 2.5 2.136 (2.564 + 2.136/3) w + 0.7 4.72

w/2

+ 0.5 3.8 4.72 w/3 0.513m 2.564 = 7 0.7m + 7 0.5m

m

wLy2

= 25.1130

Segment B:

0.5 2.136 2.5 (3 + 2.5/3) w + 0.7 32 w/2 + 0.5 5.3 5.52 w/3

+ 0.513m 3 = 6 1.2m + 7.436 0.6m

m

wLy2

= 25.2397

Segment C:

0.5 5.5 5.32 w/3 + 0.7 5.32 w/2 + 0.5 3.8 5.32 w/3

= 10 0.7m + 10 0.5m m

wLy2

= 22.4841

Segment D:

0.5 10 3.82 w/3

= 10 0.6m m

wLy2

= 24.9273

Average m

wLy2

= 24.4410

0.7

m

0.5

m

0.6 m

1.2 m

124


= 0.4, = 0.3

r = 1.0

= 0.667, k = 3.0


m

wLy2

= 28.89752

r1 = 1.8111 c1 = 5.5

r3 = 2.4148 c3 = 4.1

r4 = 2.00681 c4 = 5.0

Failure Pattern: 9


Segment A:

0.5 1.1 1.5 (4 + 1.5/3) w + 0.9 5.52 w/2

+ 0.5 5 5.52 w/3 = 7 0.7m + 7 0.5m

m

wLy2

= 19.7508

Segment B:

0.5 1.5 1.1 (3+1.1/3) w + 1.5 32 w/2 + 0.5 4.5 4.12 w/3

= 6 1.2m + 6 0.6m

m

wLy2

= 48.7915

Segment C:

0.5 4.1 4.52 w/3 + 0.9 4.52 w/2 + 0.5 5 4.52 w/3

= 10 0.7m + 10 0.5m

m

wLy2

= 30.1318

Segment D:

0.5 10 52 w/3

= (10 1.2m) m

wLy2

= 28.7977

Average m

wLy2

= 31.8915

0.5

m

0.7

m

1.2 m

0.6 m

126


= 0.3, = 0.1

r = 1.0

= 1.0, k = 3.0


m

wLy2

= 29.43841

r1 = 1.80 c1 = 5.6

r2 = 2.55 c2 = 3.9

r3 = 2.10 c3 = 4.8

Failure Pattern: 2


Q1 = 4.433/3.8 1.5m = 1.75m

Segment A:

0.5 3.8 4.433 (1.167 + 4.433/3) w + 3.8 1.1672 w/2

+ 0.5 5.2 5.62 w/3 1.75m 1.167 = 9 0.75m + 9 0.75m

m

wLy2

= 29.8652

Segment B:

0.5 4.433 3.8 (1 + 3.8/3) w + 2.6 12 w/2 + 0.5 4.82 w/2

+ 0.5 3.9 4.82 w/3 + 1.75m 1 = 8.833 1.5m

m

wLy2

= 27.9607

Segment C:

0.5 10 3.92 w/3 = (10 0.75m)

m

wLy2

= 29.5857

Segment D:

0.5 5.6 5.22 w/3 + 0.5 5.22 w/2 + 0.5 3.9 5.22 w/3

= (10 1.5m) m

wLy2

= 30.2582

Average m

wLy2

= 29.4174

0.7

5 m

0.7

5 m

1.5 m

128


= 0.3, = 0.1

r = 1.0

= 1.0, k = 3.0


m

wLy2

= 27.40698

r1 = 1.7 c1 = 5.9

r2 = 3.02857 c2 = 3.3

r3 = 2.10 c3 = 4.8

Failure Patterns: 2


Q1 = 4.671/3.8 1.5m = 1.844m

Segment A:

0.5 3.8 4.671 (1.229 + 4.671/3) w + 3.8 1.2292 w/2

+ 0.5 5.2 5.92 w/3 1.844m 1.229 = 9 1.0m + 9 0.5m

m

wLy2

= 27.2944

Segment B:

0.5 4.671 3.8 (1 + 3.8/3) w + 2.9 12 w/2 + 0.8 4.82 w/2

+ 0.5 3.3 4.82 w/3 + 1.844m 1 = 8.771 1.5m

m

wLy2

= 26.033

Segment C:

0.5 10 3.32 w/3 = (10 0.5m)

m

wLy2

= 27.5482

Segment D:

0.5 5.9 5.22 w/3 + 0.8 5.22 w/2 + 0.5 3.3 5.22 w/3

= (10 1.5m) m

wLy2

= 28.6931

Average m

wLy2

= 27.3922

1.0

m

0.5

m

1.5 m

130


= 0.3, = 0.1

r = 1.0

= 1.0, k = 3.0


m

wLy2

= 31.23402

r1 = 1.8111 c1 = 5.5

r3 = 2.1333 c3 = 4.7

r4 = 1.88238 c4 = 5.3

Failure Pattern: 9


Q1 = 4.33/3.7 1.5m = 1.755m

Segment A:

0.5 3.7 4.33 (1.17 + 4.33/3) w + 3.7 1.172 w/2

+ 0.5 5.3 5.52 w/3 1.755m 1.17 = 9 0.5m + 9 1.0m

m

wLy2

= 30.9906

Segment B:

0.5 4.33 3.7 (1 + 3.7/3) w + 2.5 12 w/2 + 0.5 4.5 4.72 w/3

+ 1.755m 1 = 8.83 1.5m

m

wLy2

= 32.1780

Segment C:

0.5 10 4.52 w/3 = (10 1.0m)

m

wLy2

= 29.6296

Segment D:

0.5 10 5.32 w/3

= (10 1.5m) m

wLy2

= 32.0398

Average m

wLy2

= 31.2095

0.5

m

1.0

m

1.5 m

132

4.30. ONE LONG EDGE CONTINUOUS SLAB (O.L.C) (Refer Fig. 4.25)

= 0.2, = 0.1

r = 1.2

= 1.0, k = 4.0


m

wLy2

= 31.95957

r1 = 2.0111 c1 = 6.0

r3 = 1.8222 c3 = 5.5

r4 = 2.31625 c4 = 4.3

Failure Pattern: 9


Q1 = 4.909/4.5 1m = 1.091m

Segment A:

0.5 4.5 4.909 (1.091 + 4.909/3) w + 4.5 1.0912 w/2 + 0.2 62

w/2

+ 0.5 4.3 62 w/3 1.091m 1.091 = 9 2.0m

m

wLy2

= 30.8514

Segment B:

0.5 4.909 4.5 (1 + 4.5/3) w + 3.6 12 w/2 + 0.5 6 5.52 w/3

+ 1.091m 1 = 9.6 1.0m + 10.909 1.0m

m

wLy2

= 32.5461

Segment C:

0.5 5.5 62 w/3 + 0.2 62 w/2 + 0.5 4.3 62 w/3

= 10 2.0m m

wLy2

= 32.0513

Segment D:

0.5 12 4.32 w/3

= 12 1.0m m

wLy2

= 32.45

Average m

wLy2

= 31.9747

2.0

m

1.0 m

1.0 m

134


= 0.2, = 0.1

r = 1.2

= 1.0, k = 4.0


m

wLy2

= 27.81781

r1 = 2.0111 c1 = 6.0

r3 = 1.7222 c3 = 5.8

r4 = 3.08465 c4 = 3.2

Failure Pattern: 9


Q1 = 4.966/4.8 0.5m = 0.517m

Segment A:

0.5 4.8 4.966 (1.034 + 4.966/3) w + 4.8 1.0342 w/2

+ 1 62 w/2 + 0.5 3.2 62 w/3 0.517m 1.034

= 9 2.0m m

wLy2

= 25.8075

Segment B:

0.5 4.966 4.8 (1 + 4.8/3) w + 3.6 12 w/2 + 0.5 6 5.82 w/3

+ 0.517m 1 = 9.6 1.5m + 10.966 0.5m

m

wLy2

= 29.15351

Segment C:

0.5 5.8 62 w/3 + 1 62 w/2 + 0.5 3.2 62 w/3

= 10 2.0m m

wLy2

= 27.7778

Segment D:

0.5 12 3.22 w/3

= 12 0.5m m

wLy2

= 29.2969

Average m

wLy2

= 28.0089

2.0

m

0.5 m

1.5 m

136


= 0.2, = 0.1

r = 1.2

= 1.0, k = 4.0


m

wLy2

= 35.50319

r1 = 2.1 c1 = 5.7

r2 = 2.10909 c2 = 5.7

r3 = 2.0 c3 = 5

Failure Pattern: 2


Q1 = 4.56/4 1.5m = 1.71m

Segment A:

0.5 4 4.56 (1.14 + 4.56/3) w + 4 1.142 w/2

+ 0.5 5 5.72 w/3 1.71m 1.14 = 9 2.0m

m

wLy2

= 36.9889

Segment B:

0.5 4.56 4 (1 + 4/3) w + 3.3 12 w/2 + 0.6 52 w/2

+ 0.5 5.7 52 w/3 + 1.171m 1 = 9.6 0.5m + 10.86 1.5m

m

wLy2

= 36.7644

Segment C:

0.5 10 5.72 w/3

= 10 2.0m m

wLy2

= 36.9344

Segment D:

0.5 5.7 52 w/3 + 0.6 52 w/2 + 0.5 5.7 52 w/3

= 12 1.5m

` m

wLy2

= 32.7272

Average m

wLy2

= 35.8537

2.0

m

1.5 m

0.5 m

138


= 0.2, = 0.3

r = 1.2

= 1.5, k = 3.0


m

wLy2

= 26.72977

r1 = 2.315 c1 = 5.2

r2 = 1.86046 c2 = 6.5

r3 = 2.01 c3 = 5

Failure Pattern: 3


Q1 = 2.692/2.8 0.9m = 0.865m

Segment A:

0.5 2.692 2.8 (2.4 + 2.8/3) w + 2 2.42 w/2

+ 0.5 5 5.22 w/3 + 0.865m 2.4

= 7 0.9m + 7.692 0.9m m

wLy2

= 27.2831

Segment B:

0.5 2.8 2.692 (2.308 + 2.692/3) w + 2.8 2.3082 w/2 + 0.3 52

w/2

+ 0.5 6.5 52 w/3 0.865m 2.308 = 9.6 1.2m

m

wLy2

= 26.8336

Segment C:

0.5 10 6.52 w/3

= 10 0.9m + 10 0.9m m

wLy2

= 25.56213

Segment D:

0.5 5.2 52 w/3 + 0.3 52 w/2 + 0.5 6.5 52 w/3

= 12 1.2m m

wLy2

= 27.4285

Average m

wLy2

= 26.7769

0.9

m

0.9

m

1.2 m

140


= 0.2, = 0.3

r = 1.2

= 1.5, k = 3.0


m

wLy2

= 26.69564

r1 = 2.315 c1 = 5.2

r2 = 1.86046 c2 = 6.5

r3 = 2.01 c3 = 5

Failure Patterns: 3


Q1 = 2.692/2.8 0.8m = 0.769m

Segment A:

0.5 2.692 2.8 (2.4 + 2.8/3) w + 2 2.42 w/2

+ 0.5 5 5.22 w/3 + 0.769m 2.4

= 7 1.0m + 7.692 0.8m m

wLy2

= 27.6777

Segment B:

0.5 2.8 2.692 (2.308 + 2.692/3) w + 2.8 2.3082 w/2 + 0.3 52

w/2

+ 0.5 6.5 52 w/3 0.769m 2.308

= 9.6 1.2m m

wLy2

= 26.3938

Segment C:

0.5 10 6.52 w/3

= 10 1.0m + 10 0.8m m

wLy2

= 25.5621

Segment D:

0.5 5.2 52 w/3 + 0.3 52 w/2 + 0.5 6.5 52 w/3

= 12 1.2m m

wLy2

= 27.4285

Average m

wLy2

= 26.7655

1.0

m

0.8

m

1.2 m

142


= 0.2, = 0.3

r = 1.2

= 1.5, k = 3.0


m

wLy2

= 26.7639

r1 = 2.315 c1 = 5.2

r2 = 1.86046 c2 = 6.5

r3 = 2.01 c3 = 5

Failure Pattern: 3


Q1 = 2.692/2.8 1m = 0.961m

Segment A:

0.5 2.692 2.8 (2.4 + 2.8/3) w + 2 2.42 w/2

+ 0.5 5 5.22 w/3 + 0.961m 2.4

= 7 0.8m + 7.692 1.0m m

wLy2

= 26.8886

Segment B:

0.5 2.8 2.692 (2.308 + 2.692/3) w + 2.8 2.3082 w/2 + 0.3 52

w/2

+ 0.5 6.5 52 w/3 0.961m 2.308 = 9.6 1.2m

m

wLy2

= 27.2735

Segment C:

0.5 10 6.52 w/3

= 10 0.8m + 10 1.0m m

wLy2

= 25.5621

Segment D:

0.5 5.2 52 w/3 + 0.3 52 w/2 + 0.5 6.5 52 w/3

= 12 1.2m m

wLy2

= 27.4286

Average m

wLy2

= 26.7882

0.8

m

1.0

m

1.2 m

144


= 0.3, = 0.2

r = 1.2

= 1.5, k = 4.0


m

wLy2

= 35.63814

r1 = 2.0111 c1 = 6

r3 = 2.31665 c3 = 4.3

r4 = 1.9595 c4 = 5.1

Failure Pattern: 9


Q1 = 3.209/2.3 0.8m = 1.116m

Segment A:

0.5 2.3 3.209 (2.791 + 3.209/3) w + 2.3 2.7912 w/2

+ 0.6 62 w/2 + 0.5 5.1 62 w/3 1.116m 2.791

= 8 2.4m m

wLy2

= 34.54013

Segment B:

0.5 3.209 2.3 (2 + 2.3/3) w + 2.4 22 w/2 + 0.5 6 4.32 w/3

+ 1.116m 2.0 = 8.4 0.8m + 9.209 0.8m

m

wLy2

= 35.3887

Segment C:

0.5 4.3 62 w/3 + 0.6 62 w/2 + 0.5 5.1 62 w/3

= 10 2.4m m

wLy2

= 35.7143

Segment D:

0.5 12 5.12 w/3

= 12 0.8m + 12 0.8m m

wLy2

= 36.9089

Average m

wLy2

= 35.638

2.4

m

0.8 m

0.8 m

146


= 0.3, = 0.2

r = 1.2

= 1.5, k = 4.0


m

wLy2

= 35.48967

r1 = 2.0111 c1 = 6

r3 = 2.21665 c3 = 4.5

r4 = 1.92193 c4 = 5.2

Failure Pattern: 9


Q1 = 3.333/2.5 0.5m = 0.667m

Segment A:

0.5 2.5 3.333 (2.667 + 3.333/3) w + 2.5 2.6672 w/2

+ 0.3 62 w/2 + 0.5 5.2 62 w/3 0.667m 2.667

= 8 2.4m m

wLy2

= 34.26176

Segment B:

0.5 3.333 2.5 (2 + 2.5/3) w + 2.4 22 w/2 + 0.5 6 4.52 w/3

+ 0.667m 2.0 = 8.4 1.1m + 9.333 0.5m

m

wLy2

= 35.1140

Segment C:

0.5 4.5 62 w/3 + 0.3 62 w/2 + 0.5 5.2 62 w/3

= 10 2.4m m

wLy2

= 37.7358

Segment D:

0.5 12 5.22 w/3

= 12 1.1m + 12 0.5m m

wLy2

= 35.5029

Average m

wLy2

= 35.4036

2.4

m

0.5 m

1.1 m

148


= 0.3, = 0.2

r = 1.2

= 1.5, k = 4.0


m

wLy2

= 35.77388

r1 = 2.0111 c1 = 6

r3 = 2.41665 c3 = 4.1

r4 = 1.90589 c4 = 5.2

Failure Patterns: 9


Q1 = 3.073/2.1 1.1m = 1.61m

Segment A:

0.5 2.1 3.073 (2.927 + 3.073/3) w + 2.1 2.9272 w/2

+ 0.7 62 w/2 + 0.5 5.2 62 w/3 1.61m 2.927

= 8 2.4m m

wLy2

= 36.4825

Segment B:

0.5 3.073 2.1 (2 + 2.1/3) w + 2.4 22 w/2 + 0.5 6 4.12 w/3

+ 1.61m 2 = 8.4 0.5m + 9.073 1.1m

m

wLy2

= 36.1464

Segment C:

0.5 4.1 62 w/3 + 0.7 62 w/2 + 0.5 5.2 62 w/3

= 10 2.4m m

wLy2

= 35.0877

Segment D:

0.5 12 5.22 w/3

= 12 0.5m + 12 1.1m m

wLy2

= 35.5029

Average m

wLy2

= 35.8049

2.4

m

1.1 m

0.5 m

150


= 0.1, = 0.3

r = 1.2

= 2.0, k = 3.0


m

wLy2

= 27.37743

r1 = 2.11111 c1 = 5.7

r3 = 2.10370 c3 = 4.8

r4 = 2.10604 c4 = 4.7

Failure Pattern: 10


Q1 = 3.789/4.5 2.0m = 1.684m

Segment A:

0.5 3.789 4.5 (1.2 + 4.5/3) w + 1.8 1.22 w/2

+ 0.5 5.72 w/2 + 0.5 4.7 5.72 w/3 + 1.684m 1.2

= 8.989 2.0m m

wLy2

= 27.5652

Segment B:

0.5 4.5 3.789 (1.011 + 3.789/3) w + 4.5 1.0112 w/2

+ 0.5 6.3 4.82 w/3 1.684m 1.011 = 10.8 1.0m

m

wLy2

= 27.2514

Segment C:

0.5 4.8 6.32 w/3 + 0.5 6.32 w/2 + 0.5 4.7 6.32 w/3

= 10 2.0m m

wLy2

= 27.4857

Segment D:

0.5 12 4.72 w/3

= 12 1.0m m

wLy2

= 27.1616

Average m

wLy2

= 27.3660

2.0

m

1.0 m

152


= 0.1, = 0.3

r = 1.2

= 2.0, k = 3.0


m

wLy2

= 29.66902

r1 = 3.13 c1 = 3.8

r2 = 2.66948 c2 = 4.5

r3 = 2.01 c3 = 5

Failure Pattern: 3


Q1 = 3.421/2.6 1.0m = 1.316m

Segment A:

0.5 3.421 2.6 (1.2 + 2.6/3) w + 2 1.22 w/2

+ 0.5 5 3.82 w/3 + 1.316m 1.2

= 8.421 1.0m m

wLy2

= 30.1874

Segment B:

0.5 2.6 3.421 (1.579 + 3.421/3) w + 2.6 1.5792 w/2 + 3.7 52

w/2

+ 0.5 52 4.5 w/3 1.316m 1.579 = 10.8 2.0m

m

wLy2

= 29.4740

Segment C:

0.5 10 4.52 w/3

= 10 1.0m m

wLy2

= 29.6296

Segment D:

0.5 3.8 52 w/3 + 3.7 52 w/2 + 0.5 4.5 52 w/3

= 12 2.0m m

wLy2

= 29.6907

Average m

wLy2

= 29.7454

1.0

m

2.0 m

Documents

010 chapter 4 - INFLIBNETshodhganga.inflibnet.ac.in/bitstream/10603/12388/10/10_chapter 4.pdf · Various support conditions of slab (Nine edge conditions). In addition to the above,