Engineering Mechanics

(^738) „„„„„ A Textbook of Engineering Mechanics
Taking moments of all the pressures about A, and equating the same,
(^) Ph× =
2
Pressure 0.5
3
ADE
⎡⎤⎛⎞
⎢⎥××⎜⎟
⎣⎦⎝⎠
0.5
Pressure 0.5
2
BFED
⎡⎤⎛⎞
+××⎢⎥⎜⎟
⎣⎦⎝⎠
2
Pressure 0.5 0.5
3
FCE
⎡⎤⎛⎞
+×+×⎢⎥⎜⎟
⎣⎦⎝⎠
6.125h =
11
4.9 0.5
23
⎡⎤⎛⎞
⎢⎥⎜⎟×× ×
⎣⎦⎝⎠
()
3
0.5 4.9
4
+××⎡⎤
⎢⎥⎣⎦^
15
9.8 0.5
26
⎡⎤⎛⎞
+×××⎢⎥⎜⎟
⎣⎦⎝⎠
= 0.408 + 1.838 + 2.042 = 4.288
∴ (^) h =
4.288
0.7 m
6.125
=^ Ans.
35.17. PRESSURE DIAGRAM DUE TO LIQUIDS ON BOTH SIDES
Consider a vertical wall subjected to pressure due to liquids on its both sides as shown in
Fig. 35.20.
Let w 1 = Specific weight of liquid 1
H 1 = Height of liquid 1, and
w 2 , H 2 = Corresponding values for the liquid 2,
We know that the pressure of liquid 1 will be zero at the
liquid surface and will increase, by a straight line law, to (w 1 H 1 )
at the bottom as shown in Fig. 35.23.
∴ Total pressure on the wall per unit length due to
liquid 1,
P 1 =
2
11
111
1
22
wH
HwH×=
Similarly, total pressure on the wall per unit length due to
liquid 2,
P 2 =
2
22
222
1
22
wH
HwH×=
A little consideration will show, that as the two pressures are acting in the opposite directions,
therefore the resultant pressure will be given by the difference of the two pressures (i.e. P = P 1 – P 2 ).
The line of action of the resultant pressure may be found out by equating the moments of P 1 and P 2
about the bottom of the wall C.
Example 35.16. A bulkhead 3 m long divides a storage tank. On one side, there is a petrol
of specific gravity 0.78 stored to a depth of 1.8 m, while on the other side there is an oil of specific
gravity 0.88 stored to a depth of 0.9 m. Determine the resultant pressure on the bulkhead, and the
position at which it acts.
Solution. Given: Length of bulkhead = 3 m.
Fig. 35.23. Pressure diagram due to
liquids on both sides.