A masonry dam (density = 20,000 N/m3) 6 m high, one metre wide at the top and 4 m wide at the base, has vertical water face. The minimum stress at the base of the dam when the reservoir is full, will be
A. 75 N/m2
B. 750 N/m2
C. 7,500 N/m2
D. 75,000 N/m2
Answer: Option C
Join The Discussion
Comments ( 16 )
A. $$\frac{2}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{8}$$
D. $$\frac{8}{5}$$
Principal planes are subjected to
A. Normal stresses only
B. Tangential stresses only
C. Normal stresses as well as tangential stresses
D. None of these
A. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
B. $$\frac{{\text{I}}}{{\text{M}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$
C. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{F}}}{{\text{Y}}}$$
D. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{Y}}}{{\text{F}}}$$
A. $$\frac{{\text{M}}}{{\text{T}}}$$
B. $$\frac{{\text{T}}}{{\text{M}}}$$
C. $$\frac{{2{\text{M}}}}{{\text{T}}}$$
D. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
Weight of dam: W1=1*6*1*20=120kN/m3; W2=0.5*3*6*1*20=180KN/m3
Hydrostatic force: F=0.5*60*6=180KN/m3.
Taking a moment at the toe of the dam: Mr= 120*3.5+180*2=780KNm; Mo=180*2=360KNm.
find X=(Mr-Mo)/ Total Vertical force=1.4m
Find, e=B/2-X=2-1.4=0.6.
Now, Min stress at base= Total vertical force/B(1-6e/B)=7.5KN/m2=7500N/m2.
Option "D" shall be correct answer as calculation reflect the answer 75,000 N/M2 [ =20,000x(LxBxH)] / [ LxB ]
So,
= [20,000x(1x{(1+4)/2}x6] / [ 1x4]
=20,000x(1x2.5x6) / 4
=3,00,000 / 4
=75,000 N/M2
Assume length is equal to 1m.
W = Density x Volume.
a=1, b=4, h=6.
W = 20000 x [1/2 x (a+b) x h x 1m],
W = 300000.
For, minimum stress = W/b (1 - 6e/b),
for stability e = b/6 = 4/6 = 0.6.
Stress = 300000/4 (1 - 6x0.6/4).
Stress 7500 N/m^2.
Thank you
Assume length is equal to 1m.
W = Density x Volume
a=1, b=4, h=6.
W = 20000 x [1/2 x (a+b) x h x 1m]
W = 300000.
For, minimum stress = W/b (1 - 6e/b)
for stability e = b/6 = 4/6 = 0.6
Stress = 300000/4 (1 - 6x0.6/4)
Stress 7500 N/m2
Assume length is equal to 1m.
W = Density x Volume
a=1, b=4, h=6.
W = 20000 x [1/2 x (a+b) x h x 1m]
W = 300000.
For, minimum stress = W/b (1 - 6e/b)
for stability e = b/6 = 4/6 = 0.6
Stress = 300000/4 (1 - 6x0.6/4)
Stress 7500 N/m2
Assume length is equal to 1m.
W = Density x Volume
a=1, b=4, h=6.
W = 20000 x [1/2 x (a+b) x h x 1m]
W = 300000.
For, minimum stress = W/b (1 - 6e/b)
for stability e = b/6 = 4/6 = 0.6
Stress = 300000/4 (1 - 6x0.6/4)
Stress 7500 N/m2
Assume length is equal to 1m.
W = Density x Volume
a=1, b=4, h=6.
W = 20000 x [1/2 x (a+b) x h x 1m]
W = 300000.
For, minimum stress = W/b (1 - 6e/b)
for stability e = b/6 = 4/6 = 0.6
Stress = 300000/4 (1 - 6x0.6/4)
Stress 7500 N/m2
Please discuss the method
Atul Kaushik,
but right ans is7500...??
How?? Please explain..
what will be correct ans?
pressure calculated only for rectangular portion
Option D is correct answer.
Density*volume/area=stress at base
(20000*(1/2*(1+4)*6))/(4*1)=75000 N/sq.m
give the solution
Solution of these question.