All Chapters Reverse
Chapter 17
17.1 Eq. (17.8)
æ f¢ ö æ 35 ö
s a = 0.65g HK a = 0.65g H tan 2 ç 45 - ÷ = (0.65)(17)(6.5) tan 2 ç 45 - ÷
è 2ø è 2ø
= 19.6 kN/m 2
19.46 kN/m2 19.46 kN/m2
1m 2m 2m 1.5 m
A B1 B2 C
SM B1 = 0
æ3ö
(19.46)(3) ç ÷
A= è 2 ø = 43.79 kN/m
2
B1 = (19.46)(3) - 43.79 = 14.59 kN/m
SM B2 = 0
æ 3.5 ö
(19.46)(3.5) ç ÷
è 2 ø
C= = 59.6 kN/m
2
B2 = (19.46)(3.5) - 59.6 = 8.51 kN/m
Strut load at A = (43.79)(spacing) = (43.79)(3) = 131.4 kN
Strut load at B = (B1 + B2)(spacing) = (14.59 + 8.51)(3) = 69.3kN
Strut load at C = (59.6)(3) = 178.8 kN
151
,17.2 a. For the sheet pile, refer to shear force diagram.
24.33 kN
1m 0.75 m
A
B′ B1
1.25 m
19.46 kN 14.49 kN
1
M A = (1)(19.46) = 9.73kN × m/m
2
29.18 kN
8.51 kN
B′′ C
B2
0.437 m 1.563 m 1.5 m
30.42 kN
1
M B¢¢ = (0.437)(8.51) = 1.86 kN × m/m
2
1
M C = (1.5)(29.18) » 21.9 kN/m
2
21.9 kN × m/m
S= = 0.129×10-3 m3 /m of wall
170 ´ 10 kN × m/m
3
Bs 2
b. For wales, M max = .
8
Bs 2 (69.38)(32 )
S= = = 0.459×10-3 m 3 /m
8s all (8)(170 ´ 10 )
3
152
, æ 40 ö
17.3 K a = ç 45 - ÷ = 0.217
è 2 ø
s a = 0.65g HK a = (0.65)(18)(6.5)(0.217) = 16.5kN/m 2
SM B1 = 0
æ3ö
(16.5)(3) ç ÷
A= è 2 ø = 37.13kN/m
2
16.5 kN/m2 16.5 kN/m2
1m 2m 2m 1.5 m
A B1 B2 C
B1 = (16.5)(3) - 37.13 = 12.37 kN/m
SM B2 = 0
æ 3.5 ö
(16.5)(3.5) ç ÷
è 2 ø
C= = 50.53 kN/m
2
B2 = (16.5)(3.5) - 50.53 = 7.22 kN/m
Strut load at A = (37.13)(4) = 148.5 kN
Strut load at B = (12.37 + 7.22)(4) = 78.4 kN
Strut load at C = (50.53)(4) = 202.12 kN
153
, 17.4 Refer to the pressure diagram in Problem 17.3. The shear force diagram is given
next.
1m 2m
20.63 kN
0.75 m
A
B1
B′
12.37 kN
16.5 kN
25.78 kN
B2
B′′ C
24.75 kN
7.22 kN
1.562 m
0.438 m 1.5 m
It can be seen that M C will be maximum.
1
M C = (1.5)(24.75) = 18.56 kN/m
2
18.56 kN × m/m 18.56
S= = = 0.109×10-3 m 3 /m of wall
s all 170 ´10 3
17.5 a. H = 8 m; H s = 3 m; H c = 5. Eq. (17.12):
1 1
g av = [g s H s + ( H - H s )g c ] = [(17.5)(3) + (5)(18.2)]
H 8
= 17.94 kN /m 3
154
Chapter 17
17.1 Eq. (17.8)
æ f¢ ö æ 35 ö
s a = 0.65g HK a = 0.65g H tan 2 ç 45 - ÷ = (0.65)(17)(6.5) tan 2 ç 45 - ÷
è 2ø è 2ø
= 19.6 kN/m 2
19.46 kN/m2 19.46 kN/m2
1m 2m 2m 1.5 m
A B1 B2 C
SM B1 = 0
æ3ö
(19.46)(3) ç ÷
A= è 2 ø = 43.79 kN/m
2
B1 = (19.46)(3) - 43.79 = 14.59 kN/m
SM B2 = 0
æ 3.5 ö
(19.46)(3.5) ç ÷
è 2 ø
C= = 59.6 kN/m
2
B2 = (19.46)(3.5) - 59.6 = 8.51 kN/m
Strut load at A = (43.79)(spacing) = (43.79)(3) = 131.4 kN
Strut load at B = (B1 + B2)(spacing) = (14.59 + 8.51)(3) = 69.3kN
Strut load at C = (59.6)(3) = 178.8 kN
151
,17.2 a. For the sheet pile, refer to shear force diagram.
24.33 kN
1m 0.75 m
A
B′ B1
1.25 m
19.46 kN 14.49 kN
1
M A = (1)(19.46) = 9.73kN × m/m
2
29.18 kN
8.51 kN
B′′ C
B2
0.437 m 1.563 m 1.5 m
30.42 kN
1
M B¢¢ = (0.437)(8.51) = 1.86 kN × m/m
2
1
M C = (1.5)(29.18) » 21.9 kN/m
2
21.9 kN × m/m
S= = 0.129×10-3 m3 /m of wall
170 ´ 10 kN × m/m
3
Bs 2
b. For wales, M max = .
8
Bs 2 (69.38)(32 )
S= = = 0.459×10-3 m 3 /m
8s all (8)(170 ´ 10 )
3
152
, æ 40 ö
17.3 K a = ç 45 - ÷ = 0.217
è 2 ø
s a = 0.65g HK a = (0.65)(18)(6.5)(0.217) = 16.5kN/m 2
SM B1 = 0
æ3ö
(16.5)(3) ç ÷
A= è 2 ø = 37.13kN/m
2
16.5 kN/m2 16.5 kN/m2
1m 2m 2m 1.5 m
A B1 B2 C
B1 = (16.5)(3) - 37.13 = 12.37 kN/m
SM B2 = 0
æ 3.5 ö
(16.5)(3.5) ç ÷
è 2 ø
C= = 50.53 kN/m
2
B2 = (16.5)(3.5) - 50.53 = 7.22 kN/m
Strut load at A = (37.13)(4) = 148.5 kN
Strut load at B = (12.37 + 7.22)(4) = 78.4 kN
Strut load at C = (50.53)(4) = 202.12 kN
153
, 17.4 Refer to the pressure diagram in Problem 17.3. The shear force diagram is given
next.
1m 2m
20.63 kN
0.75 m
A
B1
B′
12.37 kN
16.5 kN
25.78 kN
B2
B′′ C
24.75 kN
7.22 kN
1.562 m
0.438 m 1.5 m
It can be seen that M C will be maximum.
1
M C = (1.5)(24.75) = 18.56 kN/m
2
18.56 kN × m/m 18.56
S= = = 0.109×10-3 m 3 /m of wall
s all 170 ´10 3
17.5 a. H = 8 m; H s = 3 m; H c = 5. Eq. (17.12):
1 1
g av = [g s H s + ( H - H s )g c ] = [(17.5)(3) + (5)(18.2)]
H 8
= 17.94 kN /m 3
154
