blob: 69000d21c11adbeed1c471ed364741d1f910902f (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
|
//example 8.19
//calculate
//Factor of safety against overturning
//Factor of safety against sliding
//Shear friction factor
clc;funcprot(0);
//given
c=1;
H=10; //heigth of dam
hw=10; //heigth of water in reservior
wb=8.25; //bottom width
Bt=1; //top width
Hs1=0.1; //slope on upstream side
gamma_w=9.81; //unit weigth of water
gamma_m=22.4; //unit weigth of masonary
f=1400; //permissible shear stress at joint
miu=0.75; //coefficient of friction
fi=atan(0.625);
theta=atan(0.1);
W1=Bt*H*gamma_m;
W2=H*H*Hs1*gamma_m/2;
W3=H*6.25*gamma_m/2;
W4=hw*gamma_w*H*Hs1/2;
P=gamma_w*hw^2/2;
U=wb*gamma_w*hw*c/2;
SumV=W1+W2+W3+W4-U;
L3=2*(wb-(Hs1*H)-Bt)/3;
L1=(wb-(Hs1*H)-Bt)+Bt/2;
L2=(wb-(Hs1*H)-Bt)+Bt+(Hs1*H/3);
L4=(wb-(Hs1*H)-Bt)+Bt+(2*Hs1*H/3);
L5=2*wb/3;L6=hw/3;
M1=W1*L1;M2=W2*L2;M3=W3*L3;M4=W4*L4;
M5=U*L5;M6=P*L6;
SumM=M1+M2+M3+M4-M5-M6;
Mplus=M1+M2+M3+M4;
Mminus=M5+M6;
FOS=miu*SumV/P;
SFF=(miu*SumV+wb*1400)/P;
FOO=Mplus/Mminus;
FOS=round(FOS*100)/100;
SFF=round(SFF*10)/10;
FOO=round(FOO*100)/100;
mprintf("Factor of safety against sliding=%f.",FOS);
mprintf("\nShear friction factor=%f.",SFF);
mprintf("\nFactor of safety against overturning=%f.",FOO);
mprintf("\nDam is unsafe against overturning");
|
