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3 (75 mm) -3 75 (175 mm) (125 mm) 150 mm 75 mm (100 mm) 1100 mm) -22 (ww 095) - 16 wwood) - 18- (450 mm) -12 - (300 mm) 4
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Answer #1

Solution:

FOR FIGURE (A)

For # 8 bar, diameter = 1 inch

so, area of 4 # 8 bars = 4 * (\pi/4) * 12 = 3.14 in2

3 6 3 E = 0.003 - - - - --- 4 + ((a-472) 4# 8 TL & = 0.03(d-c)/C T=ASY -17-

Now consider strain diagram as shown abovw,

By rule of similar triangle,

\frac{0.003}{C} = \frac{\varepsilon _{t}}{(d - C)}

\therefore  \epsilon _{t}= \frac{0.03 * (d - C)}{C}

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, \epsilon _{t}= \frac{0.03 * (d - C)}{C}

\epsilon _{t}= 0.03472 which falls in transition zone as 0.002 < \epsilon _{t} < 0.005

Hence \phi = 0.9

Design moment = \phi *  Mst = 0.9 * 3694.5 = 3325 kilo pounds inch

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