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Solutions to Chapter 19 Problems
SOLUTIONS TO CHAPTER 19 PROBLEMS S.19.1
Referring to Fig. S.19.1 1 2.5 A 1 2.5
B
B
1 2.5
5m
A
45
1 2.5 10 m
FIGURE S.19.1
the work absorbed in the yield lines is: for the two panels A, 1 2m 10 ¼ 8 m 2:5 for the two panels B,
1 25 ð0:4 m þ 0:6 mÞ ¼ 4 m 2:5
1 1 The work done by the load ¼ 14 5 5 þ55 ¼ 291:7 kNm 2 3 Then 12 m ¼ 291:7 so that m ¼ 24:31 kNm=m: S.19.2
Consider the slab shown in Fig. S.19.2(a). The work absorbed in the yield lines is: for panel A, 1 m 10 ¼ 3:33 m 3
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Solutions to Chapter 19 Problems 1 3
45 1 3
1 3 A
3m 1 3
C B
3m
C
4m
4m
B
1 4 (a)
A
45
1 4
10 m
(b)
7m
3m
FIGURE S.19.2
for panel B, 2 m 10
1 ¼ 5:0 m 4
for panel C, 1 0:6 m 7 ¼ 1:4 m 3 The total work absorbed by the yield lines is therefore 9.73 m. 1 1 þ77 ¼ 315 kNm The work done by the load ¼ 10 7 3 3 2 Therefore 9:73 m ¼ 315 i.e. m ¼ 32:37 kNm=m For the slab shown in Fig. S.19.2(b) the work absorbed in the yield lines is: for panel A, 1 ¼ 2:33 m m7 3 for panel B, 1 ¼ 3:5 m 2m7 4
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Solutions Manual for panel C, 1 ¼ 1:4 m 0:6 m 7 3 The total work absorbed in the yield lines is therefore 7.23 m. 1 1 1 þ74 þ33 ¼ 255 kNm: The work done by the load ¼ 10 7 3 3 2 2 Therefore 7:23 m ¼ 255 i.e. m ¼ 35:27 kNm=m:
S.19.3
Referring to Fig. S.19.3 x
1 2.5 B
1 x
(12 kN/m2)
(8 kN/m2)
2.5 m
A B
2.5 m 1 2.5 7m
3m
FIGURE S.19.3
the work absorbed in the yield lines is: for panel A, 1 2m ¼ 0:4 m 5 x x for panel B, 1 ¼ 16 m 2ðm þ mÞ 10 2:5 The total work absorbed by the yield lines is therefore m[16 þ (2/x)].
Solutions to Chapter 19 Problems 1 1 1 þ 5ð7 xÞ þ8 35 The work done by the load is ¼ 12 5x 3 2 2 ¼ 270 10x Therefore
2 ¼ 270 10x m 16 þ x i.e. 270 10x m¼ 2 16 þ x Differentiating the above with respect to x and simplifying gives 8x 2 þ 2x 27 ¼ 0 which gives x ¼ 1:716 m Then m ¼ 14:73 kNm=m: S.19.4
Referring to Fig. S.19.4 x
1 x
1 2.5 A
2.5 m
A
2.5 m
B
1 2.5 8m
FIGURE S.19.4
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Solutions Manual the work absorbed in the yield lines is: for panels A, 2ð20 þ 16Þ 8
1 ¼ 230:4 2:5
for panel B, 1 120 ¼ ð16 þ 8Þ 5 x x 1 1 2:5x The work done by the load ¼ w ð8 xÞ 5 þ 5x ¼ w 20 2 3 3 Then 2:5x 120 ¼ 230:4 þ w 20 3 x so that 120 230:4 þ x w¼ 2:5x 20 3 For a minimum value of w, dw/dx 5 0. Differentiating the above with respect to x and simplifying gives x 2 þ 1:0417x 12:5 ¼ 0 which gives x ¼ 3:053 m Substituting for x in the above expression for w gives w ¼ 15:45 kN=m2 : S.19.5
Referring to Fig. S.19.5 the work absorbed in the yield lines is: for panels A, 2ðm 5 þ 1:2 m 5Þ
1 ¼ 8:8 m 2:5
for panels B, 1 ¼ 5:0 m 2ðm 2:5Þ 1
Solutions to Chapter 19 Problems 2.5 m
1.0 m
2.5 m
1 1
1 1 B
B
1 2.5
A
1.0 m
A
1 2.5
(4 x) m
x
C 1 x
FIGURE S.19.5
for panel C, ðm 6 þ 1:2 m 6Þ
1 13:2 m ¼ x x
The total work absorbed by the yield lines is then ¼ m[13.8 þ (13.2/x)].
1 1 þ 2ð4 x Þ 2:5 The work done by the load ¼ 15 2 2:5 1 3 2 1 1 1 þ 1x þ 1ð4 xÞ ¼ 235 20x þ2 2:5x 3 2 1 Then 13:2 m 13:8 þ ¼ 235 20x x For x ¼ 2:0 m, m ¼ 9:56 kNm=m For x ¼ 2:5 m, m ¼ 9:70 kNm=m For x ¼ 3:0 m, m ¼ 9:62 kNm=m i.e. m ¼ 9:70 kNm=m:
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Solutions Manual
S.19.6
Referring to Fig. S.19.6 2.5 m 1 x B
8m
1 2.5
x
A
x B
A
8m
1 x
2.5 m
1 2.5
FIGURE S.19.6
the work absorbed in the yield lines is: for the panels A,
1 ¼ mð10:96 þ 0:8xÞ 2½mð2:5 þ xÞ þ ð1:4 m 8Þ 2:5 for the panels B, 1 12 m ¼ 2ðm 2:5 þ 1:4 m 2:5Þ x x 12 Therefore the total work absorbed in the yield lines ¼ m 10:96 þ 0:8x þ x 1 1 1 þ 2:5x 2 þ 2 2:5ð5:5 xÞ The work done by the applied load ¼ 20 2:52 3 3 2 ¼ 316:67 16:67x Therefore 12 ¼ 316:67 16:67x m 10:96 þ 0:8x þ x
Solutions to Chapter 19 Problems
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so that 316:67 16:67x m¼ 12 10:96 þ 0:8x þ x The maximum value of m is most easily found using the trial and error method. Then, For x ¼ 2:0 m, m ¼ 15:27 kNm=m For x ¼ 2:5 m, m ¼ 15:48 kNm=m For x ¼ 3:0 m, m ¼ 15:36 kNm=m Therefore the required value of the moment parameter is 15.48 kNm/m. S.19.7
The solution for the first part of the problem is identical to that for P.19.1 and gives m ¼ 24:31 kNm=m: Consider the secondary yield line pattern shown in Fig. S.19.7. 1/2.5
A B
B A
L
10-2L
5m 1/2.5
L
FIGURE S.19.7
For each of the panels A the work absorbed by the yield lines is m(102 L)(1/2.5) and for each of the panels B the work absorbed by the yield lines is 0.4 m 5 (1/2.5). Also the work done by the load is w[5 5 (1/3) þ 5(52 L)(1/2)]. Then, equating the work absorbed by the yield lines to the work done by the load 2 mð10 2 LÞð1=2:5Þ þ 2 0:4 m 5 ð1=2:5Þ ¼ w½5 5 ð1=3Þ þ 5ð5 2 LÞð1=2Þ Simplifying and substituting w ¼ 14 kN/m2 gives 2 mð6 LÞ ¼
17:5 ð125 30 LÞ 6
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Solutions Manual from which L ¼ 1:875 m
S.19.8
The yield line pattern for the top half of the slab is shown in Fig. S.19.8(a).
2m
3m
2m
1/2
A 1/2
2m C
C
1/2 3m
B
(a)
1/3 1/3.5
B 1/3.5
1/3.5 E
F A
A
B
(b)
45°
1/3.5
FIGURE S.19.8
Then the work absorbed by the yield lines is: Panel A: m 7 (1/2) ¼ 3.5 m Panel B: m 7 (1/3) þ 1.5 m 7 (1/3) ¼ 5.83 m Panels C: 2 m 5 (1/2) ¼ 5.0 m The total work absorbed by the panels is then 14.33 m. The work done by the load ¼ 15[5 4 (1/3) þ 5 3 (1/2)] ¼ 212.55
Solutions to Chapter 19 Problems
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Equating the two values gives m ¼ 14.83 kNm/m. Now consider the yield line pattern shown in Fig. S.19.8(b) and let the moment of resistance of the beam EF be Mb. The work absorbed by the yield lines is then Panels A: 2[m 10 (1/3.5) þ Mb (1/3.5)] ¼ 5.714 m þ 0.571 Mb Panels B: 2 m 7 (1/3.5) ¼ 4.0 m The total work absorbed by the yield lines is then 9.714 m þ 0.571 Mb. The work done by the load ¼ 15[7 7 (1/3) þ 7 3 (1/2)] ¼ 402.45 Equating the two values gives 9:714 14:83 þ 0:571 Mb ¼ 402:45 from which Mb ¼ 452:5 kNm S.19.9
Referring to Fig. S.19.9 the work absorbed by the yield lines is as follows. 2.2 m
A C B
D
a
1/2.8 a D
B
1/2.2
5.6 m
C 1/2.8 I
2.2 m
A 1/2.2
5m
4m
5m
FIGURE S.19.9
Panels A: 2 m 14 (1/2.2) ¼ 12.73 m Panels B: 2 m 10 (1/2.2) ¼ 9.09 m Panels C: 2 m 9.6 (1/2.8) þ 2 1.2 m 9.6 (1/2.8) ¼ 15.09 m Panels D: 2 m 5.6 (1/2.8) þ 2 1.2 m 5.6 (1/2.8) ¼ 8.80 m The total work absorbed by the slab is therefore 45.71 m. The work done by the load is given by Work done ¼ 14½4 ð2:2Þ2 ð1=3Þ þ 2 2:8 5:6 ð2=3Þ þ 2 9:6 2:2 ð1=2Þ þ2 5:6 2:2 ð1=2Þ þ 2 4 2:8 ð1=2Þ ¼ 1008
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Solutions Manual Note that this is obtained by dividing the slab into a combination of rectangles, squares and triangles, calculating their areas and multiplying by the displacement of their centroids. Therefore 45:71 m ¼ 1008 so that m ¼ 22:05 kNm=m
S.19.10 Consider Panel I shown in Fig. S.19.10(a). 1 xL A
L
1 0.5 L
B
A
xL
1
xL
1 xL (a)
0.5 L
FIGURE S.19.10 (a)
Suppose that the collapse load intensity is n. The work absorbed in the yield lines is Work ðyield linesÞ ¼ 2ð1:5M þ M Þ0:5Lð1=xLÞ þ ð1:5ML þ M 2LÞð1=0:5LÞ i.e. Work ðyield linesÞ ¼ ð3 þ 4x þ 2:5=x ÞM
(i)
The work done by the collapse load is Work ðcollapse loadÞ ¼ n½0:5L 2xL=3 þ 0:5LðL 2xLÞ=2 i.e. Work ðcollapse loadÞ ¼ nL2 ðx=3 þ 1=4 x=2Þ
(ii)
Solutions to Chapter 19 Problems
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Equating Eqs (i) and (ii) and simplifying nL2 3 þ 4x þ 2:5=x 12ð3x þ 4x 2 þ 2:5Þ ¼ ¼ M 1=4 x=6 3x 2x 2 For the minimum collapse load dn/dx ¼ 0. Therefore, differentiating Eq. (iii), equating to zero and simplifying 3x þ 4x 2 þ 2:5 3 þ 8x ¼ 3x 2x 2 3 4x Simplifying gives a quadratic equation in x, i.e. 18x 2 þ 10x 7:5 ¼ 0 the solution of which is x ¼ 0.425. Substituting this value of x in Eq. (iii) gives n ¼ 59:12M =L2 For Panel II, shown in Fig. S.19.10 (b), suppose the sagging moment is m0 . 2 L
1 2 L
2 L
2 L (b)
L
FIGURE S.19.10 (b)
Then 4ð1:5M þ m0 ÞL 2=L ¼ nL2 =3 Substituting the value of n gives m0 ¼ 0.964M.
(iii)
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S.19.11 (a) Possible yield line mechanisms of collapse are shown in Fig. S.19.11(a). (b) Referring to Fig. S.19.11(b), the work absorbed in the yield lines is given by
i.e.
Work ðyield linesÞ ¼ 1:2M 8 ð1=3:6Þ þ 1:2M 5 ð1=1:5Þ þ M 5:1 ð1=x Þ þ M 3:6 ð1=3Þ Work ðyield linesÞ ¼ M ð7:87 þ 5:1=x Þ
(i)
The work done by the load is Work ðloadÞ ¼ 14½ð5:1x=3Þ þ ð3:6 3=3Þ þ 5:1ð5 x Þ=2
(a)
FIGURE S.19.11 (a)
I 1.5 1.5 m
B D I 3
A
FIGURE S.19.11 (b)
I x
I 3.6
3m (b)
C
(5 – x) m
xm
3.6 m
Solutions to Chapter 19 Problems
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i.e. Work ðloadÞ ¼ 14ð16:35 0:85x Þ
(ii)
Equating Eqs (i) and (ii) and rearranging M 16:35 0:85x 2 ¼ 7:87x þ 5:1 14
(iii)
For a minimum value of M, dM/dx ¼ 0. Therefore, differentiating Eq. (iii), equating to zero and simplifying gives x 2 þ 1:3x 12:5 ¼ 0 the solution of which is x ¼ 2.94 m. Substituting this value of x in Eq. (iii) gives M ¼ 20:19 kNm=m:
Referring to Fig. S.19.1 1 2.5 A 1 2.5
B
B
1 2.5
5m
A
45
1 2.5 10 m
FIGURE S.19.1
the work absorbed in the yield lines is: for the two panels A, 1 2m 10 ¼ 8 m 2:5 for the two panels B,
1 25 ð0:4 m þ 0:6 mÞ ¼ 4 m 2:5
1 1 The work done by the load ¼ 14 5 5 þ55 ¼ 291:7 kNm 2 3 Then 12 m ¼ 291:7 so that m ¼ 24:31 kNm=m: S.19.2
Consider the slab shown in Fig. S.19.2(a). The work absorbed in the yield lines is: for panel A, 1 m 10 ¼ 3:33 m 3
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Solutions to Chapter 19 Problems 1 3
45 1 3
1 3 A
3m 1 3
C B
3m
C
4m
4m
B
1 4 (a)
A
45
1 4
10 m
(b)
7m
3m
FIGURE S.19.2
for panel B, 2 m 10
1 ¼ 5:0 m 4
for panel C, 1 0:6 m 7 ¼ 1:4 m 3 The total work absorbed by the yield lines is therefore 9.73 m. 1 1 þ77 ¼ 315 kNm The work done by the load ¼ 10 7 3 3 2 Therefore 9:73 m ¼ 315 i.e. m ¼ 32:37 kNm=m For the slab shown in Fig. S.19.2(b) the work absorbed in the yield lines is: for panel A, 1 ¼ 2:33 m m7 3 for panel B, 1 ¼ 3:5 m 2m7 4
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Solutions Manual for panel C, 1 ¼ 1:4 m 0:6 m 7 3 The total work absorbed in the yield lines is therefore 7.23 m. 1 1 1 þ74 þ33 ¼ 255 kNm: The work done by the load ¼ 10 7 3 3 2 2 Therefore 7:23 m ¼ 255 i.e. m ¼ 35:27 kNm=m:
S.19.3
Referring to Fig. S.19.3 x
1 2.5 B
1 x
(12 kN/m2)
(8 kN/m2)
2.5 m
A B
2.5 m 1 2.5 7m
3m
FIGURE S.19.3
the work absorbed in the yield lines is: for panel A, 1 2m ¼ 0:4 m 5 x x for panel B, 1 ¼ 16 m 2ðm þ mÞ 10 2:5 The total work absorbed by the yield lines is therefore m[16 þ (2/x)].
Solutions to Chapter 19 Problems 1 1 1 þ 5ð7 xÞ þ8 35 The work done by the load is ¼ 12 5x 3 2 2 ¼ 270 10x Therefore
2 ¼ 270 10x m 16 þ x i.e. 270 10x m¼ 2 16 þ x Differentiating the above with respect to x and simplifying gives 8x 2 þ 2x 27 ¼ 0 which gives x ¼ 1:716 m Then m ¼ 14:73 kNm=m: S.19.4
Referring to Fig. S.19.4 x
1 x
1 2.5 A
2.5 m
A
2.5 m
B
1 2.5 8m
FIGURE S.19.4
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Solutions Manual the work absorbed in the yield lines is: for panels A, 2ð20 þ 16Þ 8
1 ¼ 230:4 2:5
for panel B, 1 120 ¼ ð16 þ 8Þ 5 x x 1 1 2:5x The work done by the load ¼ w ð8 xÞ 5 þ 5x ¼ w 20 2 3 3 Then 2:5x 120 ¼ 230:4 þ w 20 3 x so that 120 230:4 þ x w¼ 2:5x 20 3 For a minimum value of w, dw/dx 5 0. Differentiating the above with respect to x and simplifying gives x 2 þ 1:0417x 12:5 ¼ 0 which gives x ¼ 3:053 m Substituting for x in the above expression for w gives w ¼ 15:45 kN=m2 : S.19.5
Referring to Fig. S.19.5 the work absorbed in the yield lines is: for panels A, 2ðm 5 þ 1:2 m 5Þ
1 ¼ 8:8 m 2:5
for panels B, 1 ¼ 5:0 m 2ðm 2:5Þ 1
Solutions to Chapter 19 Problems 2.5 m
1.0 m
2.5 m
1 1
1 1 B
B
1 2.5
A
1.0 m
A
1 2.5
(4 x) m
x
C 1 x
FIGURE S.19.5
for panel C, ðm 6 þ 1:2 m 6Þ
1 13:2 m ¼ x x
The total work absorbed by the yield lines is then ¼ m[13.8 þ (13.2/x)].
1 1 þ 2ð4 x Þ 2:5 The work done by the load ¼ 15 2 2:5 1 3 2 1 1 1 þ 1x þ 1ð4 xÞ ¼ 235 20x þ2 2:5x 3 2 1 Then 13:2 m 13:8 þ ¼ 235 20x x For x ¼ 2:0 m, m ¼ 9:56 kNm=m For x ¼ 2:5 m, m ¼ 9:70 kNm=m For x ¼ 3:0 m, m ¼ 9:62 kNm=m i.e. m ¼ 9:70 kNm=m:
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Solutions Manual
S.19.6
Referring to Fig. S.19.6 2.5 m 1 x B
8m
1 2.5
x
A
x B
A
8m
1 x
2.5 m
1 2.5
FIGURE S.19.6
the work absorbed in the yield lines is: for the panels A,
1 ¼ mð10:96 þ 0:8xÞ 2½mð2:5 þ xÞ þ ð1:4 m 8Þ 2:5 for the panels B, 1 12 m ¼ 2ðm 2:5 þ 1:4 m 2:5Þ x x 12 Therefore the total work absorbed in the yield lines ¼ m 10:96 þ 0:8x þ x 1 1 1 þ 2:5x 2 þ 2 2:5ð5:5 xÞ The work done by the applied load ¼ 20 2:52 3 3 2 ¼ 316:67 16:67x Therefore 12 ¼ 316:67 16:67x m 10:96 þ 0:8x þ x
Solutions to Chapter 19 Problems
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so that 316:67 16:67x m¼ 12 10:96 þ 0:8x þ x The maximum value of m is most easily found using the trial and error method. Then, For x ¼ 2:0 m, m ¼ 15:27 kNm=m For x ¼ 2:5 m, m ¼ 15:48 kNm=m For x ¼ 3:0 m, m ¼ 15:36 kNm=m Therefore the required value of the moment parameter is 15.48 kNm/m. S.19.7
The solution for the first part of the problem is identical to that for P.19.1 and gives m ¼ 24:31 kNm=m: Consider the secondary yield line pattern shown in Fig. S.19.7. 1/2.5
A B
B A
L
10-2L
5m 1/2.5
L
FIGURE S.19.7
For each of the panels A the work absorbed by the yield lines is m(102 L)(1/2.5) and for each of the panels B the work absorbed by the yield lines is 0.4 m 5 (1/2.5). Also the work done by the load is w[5 5 (1/3) þ 5(52 L)(1/2)]. Then, equating the work absorbed by the yield lines to the work done by the load 2 mð10 2 LÞð1=2:5Þ þ 2 0:4 m 5 ð1=2:5Þ ¼ w½5 5 ð1=3Þ þ 5ð5 2 LÞð1=2Þ Simplifying and substituting w ¼ 14 kN/m2 gives 2 mð6 LÞ ¼
17:5 ð125 30 LÞ 6
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Solutions Manual from which L ¼ 1:875 m
S.19.8
The yield line pattern for the top half of the slab is shown in Fig. S.19.8(a).
2m
3m
2m
1/2
A 1/2
2m C
C
1/2 3m
B
(a)
1/3 1/3.5
B 1/3.5
1/3.5 E
F A
A
B
(b)
45°
1/3.5
FIGURE S.19.8
Then the work absorbed by the yield lines is: Panel A: m 7 (1/2) ¼ 3.5 m Panel B: m 7 (1/3) þ 1.5 m 7 (1/3) ¼ 5.83 m Panels C: 2 m 5 (1/2) ¼ 5.0 m The total work absorbed by the panels is then 14.33 m. The work done by the load ¼ 15[5 4 (1/3) þ 5 3 (1/2)] ¼ 212.55
Solutions to Chapter 19 Problems
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Equating the two values gives m ¼ 14.83 kNm/m. Now consider the yield line pattern shown in Fig. S.19.8(b) and let the moment of resistance of the beam EF be Mb. The work absorbed by the yield lines is then Panels A: 2[m 10 (1/3.5) þ Mb (1/3.5)] ¼ 5.714 m þ 0.571 Mb Panels B: 2 m 7 (1/3.5) ¼ 4.0 m The total work absorbed by the yield lines is then 9.714 m þ 0.571 Mb. The work done by the load ¼ 15[7 7 (1/3) þ 7 3 (1/2)] ¼ 402.45 Equating the two values gives 9:714 14:83 þ 0:571 Mb ¼ 402:45 from which Mb ¼ 452:5 kNm S.19.9
Referring to Fig. S.19.9 the work absorbed by the yield lines is as follows. 2.2 m
A C B
D
a
1/2.8 a D
B
1/2.2
5.6 m
C 1/2.8 I
2.2 m
A 1/2.2
5m
4m
5m
FIGURE S.19.9
Panels A: 2 m 14 (1/2.2) ¼ 12.73 m Panels B: 2 m 10 (1/2.2) ¼ 9.09 m Panels C: 2 m 9.6 (1/2.8) þ 2 1.2 m 9.6 (1/2.8) ¼ 15.09 m Panels D: 2 m 5.6 (1/2.8) þ 2 1.2 m 5.6 (1/2.8) ¼ 8.80 m The total work absorbed by the slab is therefore 45.71 m. The work done by the load is given by Work done ¼ 14½4 ð2:2Þ2 ð1=3Þ þ 2 2:8 5:6 ð2=3Þ þ 2 9:6 2:2 ð1=2Þ þ2 5:6 2:2 ð1=2Þ þ 2 4 2:8 ð1=2Þ ¼ 1008
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Solutions Manual Note that this is obtained by dividing the slab into a combination of rectangles, squares and triangles, calculating their areas and multiplying by the displacement of their centroids. Therefore 45:71 m ¼ 1008 so that m ¼ 22:05 kNm=m
S.19.10 Consider Panel I shown in Fig. S.19.10(a). 1 xL A
L
1 0.5 L
B
A
xL
1
xL
1 xL (a)
0.5 L
FIGURE S.19.10 (a)
Suppose that the collapse load intensity is n. The work absorbed in the yield lines is Work ðyield linesÞ ¼ 2ð1:5M þ M Þ0:5Lð1=xLÞ þ ð1:5ML þ M 2LÞð1=0:5LÞ i.e. Work ðyield linesÞ ¼ ð3 þ 4x þ 2:5=x ÞM
(i)
The work done by the collapse load is Work ðcollapse loadÞ ¼ n½0:5L 2xL=3 þ 0:5LðL 2xLÞ=2 i.e. Work ðcollapse loadÞ ¼ nL2 ðx=3 þ 1=4 x=2Þ
(ii)
Solutions to Chapter 19 Problems
e403
Equating Eqs (i) and (ii) and simplifying nL2 3 þ 4x þ 2:5=x 12ð3x þ 4x 2 þ 2:5Þ ¼ ¼ M 1=4 x=6 3x 2x 2 For the minimum collapse load dn/dx ¼ 0. Therefore, differentiating Eq. (iii), equating to zero and simplifying 3x þ 4x 2 þ 2:5 3 þ 8x ¼ 3x 2x 2 3 4x Simplifying gives a quadratic equation in x, i.e. 18x 2 þ 10x 7:5 ¼ 0 the solution of which is x ¼ 0.425. Substituting this value of x in Eq. (iii) gives n ¼ 59:12M =L2 For Panel II, shown in Fig. S.19.10 (b), suppose the sagging moment is m0 . 2 L
1 2 L
2 L
2 L (b)
L
FIGURE S.19.10 (b)
Then 4ð1:5M þ m0 ÞL 2=L ¼ nL2 =3 Substituting the value of n gives m0 ¼ 0.964M.
(iii)
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Solutions Manual
S.19.11 (a) Possible yield line mechanisms of collapse are shown in Fig. S.19.11(a). (b) Referring to Fig. S.19.11(b), the work absorbed in the yield lines is given by
i.e.
Work ðyield linesÞ ¼ 1:2M 8 ð1=3:6Þ þ 1:2M 5 ð1=1:5Þ þ M 5:1 ð1=x Þ þ M 3:6 ð1=3Þ Work ðyield linesÞ ¼ M ð7:87 þ 5:1=x Þ
(i)
The work done by the load is Work ðloadÞ ¼ 14½ð5:1x=3Þ þ ð3:6 3=3Þ þ 5:1ð5 x Þ=2
(a)
FIGURE S.19.11 (a)
I 1.5 1.5 m
B D I 3
A
FIGURE S.19.11 (b)
I x
I 3.6
3m (b)
C
(5 – x) m
xm
3.6 m
Solutions to Chapter 19 Problems
e405
i.e. Work ðloadÞ ¼ 14ð16:35 0:85x Þ
(ii)
Equating Eqs (i) and (ii) and rearranging M 16:35 0:85x 2 ¼ 7:87x þ 5:1 14
(iii)
For a minimum value of M, dM/dx ¼ 0. Therefore, differentiating Eq. (iii), equating to zero and simplifying gives x 2 þ 1:3x 12:5 ¼ 0 the solution of which is x ¼ 2.94 m. Substituting this value of x in Eq. (iii) gives M ¼ 20:19 kNm=m: