A gravity dam of trapezoidal cross- section with one face vertical and horizontal base is 22 m. high and has a thickness of 4 m. at the top. Water upstream stands 2 m. below the crest of the dam. The specific gravity of masonry is 2.4. A. Neglecting hydrostatic uplift 1. Find the base width B of the dam so that the resultant force will cut the extreme edge of the middle third near the toe. 2. Compute the factors of safety against sliding and overturning. Use H=0.5

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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PLEASE ANSWER ONLY NUMBER 2 AND Use μ=0.6

 

THIS IS THE FORMULA BUT I COULDNT GET THE ANSWER BECAUSE I DON'T KNOW THE VALUE OF GAMMA (?). I JUST REPLACE THE VALUE OF μ OF .6

???=???/??

=0.6[24(11.175)( ?)+115.2 ? / 200? 

PS: AND MAYBE YOU CAN TEACH ME WHAT IS THE VALUE OF GAMMA IN HERE BECUASE I DON'T KNOW. IF I KNOW THE VALUE OF IT I WOULD NOT ASK HERE.

A gravity dam of trapezoidal cross-
section with one face vertical and
horizontal base is 22 m. high and has
a thickness of 4 m. at the top. Water
upstream stands 2 m. below the crest
of the dam. The specific gravity of
masonry is 2.4.
A. Neglecting hydrostatic uplift
1. Find the base width B of the
dam so that the resultant force will cut
the extreme edge of the middle third
near the toe.
2. Compute the factors of safety
against sliding and overturning. Use
H=0.5
Transcribed Image Text:A gravity dam of trapezoidal cross- section with one face vertical and horizontal base is 22 m. high and has a thickness of 4 m. at the top. Water upstream stands 2 m. below the crest of the dam. The specific gravity of masonry is 2.4. A. Neglecting hydrostatic uplift 1. Find the base width B of the dam so that the resultant force will cut the extreme edge of the middle third near the toe. 2. Compute the factors of safety against sliding and overturning. Use H=0.5
Factors of safety:
Factor of safety against sliding:
FSS
μRy
Rx
0.5[24(11.175) (y) + 115.2 y
200y
FSS = 0.9585
Factor of safety against
overturning:
RM
FSO
=
OM
16(11.175)²y+83.2(11.175)y-166.4y
1333.33y
FS。
= 2.07
Transcribed Image Text:Factors of safety: Factor of safety against sliding: FSS μRy Rx 0.5[24(11.175) (y) + 115.2 y 200y FSS = 0.9585 Factor of safety against overturning: RM FSO = OM 16(11.175)²y+83.2(11.175)y-166.4y 1333.33y FS。 = 2.07
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