HW_8 Solutions

HW_8 Solutions -...

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8-12. W 4 3m 29.8m Assumptions: KN KN > Specific weight of the concrete dam and water are = 23.6—--andy 9.8—3--. m m > Friction coefficient p = 0.65 > There is a gate on top of the dam and weight of water on top of the dam is W 4 > Consider unit width of the dam We are going to check the followings for this problem: a) Factor of safety against overturning b) Factor of safety against sliding c) Check shear stress of concrete at critical section of the dam d) Check compressive stress at the base of the dam a) Factor of safety against overturning W1=O.5*19*19!21*23.6=2129.9 KN W2=3*26*1*23 6=1840 8 KN W3=0.5*26*26t1.5,1*23.6=53 17.9 KN W4=3* 1*1*9.8=29.4 KN (negligible compared to others) W=W1 +W2+W3+W4=9288.6 KN This acts at:
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* 23.5+ *18.8+ W *11.6 =15.8 m from point o (W 4 is small so not considered) x 1 = w Now calculate sum of horizontal forces First I am going to prove a simple formula for finding the center of mass of trapezoid and I am going to use it through out my solution to find out where the resultant forces act for trapezoidal hydrostatic forces. T B H x T * H * + (B - T) * * 3 B+2TH - B+T 3 / x= Now calculate magnitude of forces one by one along with the location of their influence:
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F =(ly +8y 1 )*-*1=9*9.8* -=308.7KN .
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