Solution:
FOR FIGURE (A)
For # 8 bar, diameter = 1 inch
so, area of 4 # 8 bars = 4 * (/4)
* 12 = 3.14 in2
Now consider strain diagram as shown abovw,
By rule of similar triangle,
Let 'a' is the depth of compression zone.
Let C'1 is compression force acting at a distance = 4 / 2 = 2" from extreme top.
Let C''1 is compression force acting at a distance = 4 + ((a-4)/2) from extreme top.
Let T is the total tensile force acting at the depth equal to 20" from top = As * fy = 3.14 * 60 = 188.4 Kilo pounds.
Now this total T has to be resisted by total compression force ( C'1 + C''1 ) so as to have equilibrium
hence T = C'1 + C''1
240 = 0.85 * f'c * [ 2* (3 * 4 ) + 12* (a -4) ]
240 = 0.85 * 4 * [ 2* (3 * 4 ) + 12* (a -4) ]
a= 7.88 inch
Now to calculate C, C = a / 0.85 = 9.27 inch
Hence, C'1 = 0.85 * 4 * ( 2* (3 * 4 )) = 81.6 Kilo pound
C''1 = 0.85 * 4 * 12* (a -4)
= 0.85 * 4 * 12* (7.88 -4) = 158.3 Kilo pound
Now taking moment of the above two forces about the tensile force, we get
Mst =( C'1 * (d-2)) + ( C''1 * (d-(2+a/2)))
Mst =( 81.6 * (20-2)) + ( 158.3 * (20-(2+7.88/2)))
Mst = 1468.8 + 2225.7 = 3694.5 kilo pounds inch
Now,
0.03472 which falls in transition zone as 0.002 <
< 0.005
Hence
= 0.9
Design moment =
* Mst = 0.9 * 3694.5 = 3325 kilo
pounds inch
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