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TitleExample Problems
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAÑETE

EXAMPLES
Copyright © 2013





The db is the diameter of the bar which is 28 mm. Substitute it
And we can get the development length.

Ld = 54.69
db

Ld = 54.69 (28)

Then, the value of development length will be:
Ld = 1531.29 mm


b.) Nominal moment capacity:
At first we must compute for the area in order to compute the moment capacity.

As = π (28)2 (3) = 1847 mm2
4
T=Asfy
And the, we can now compute for the value of a.
T = 0.85f’cab
Asfy = 0.85f’cab
1847 (414.7) = 0.85(20.7) (a) (400)
a = 108.83 mm
By getting the unknown value, we can now substitute it to compute for the moment capacity of the
reinforced concrete.
Mn = Asfy (d – a/2)
Mn = 1847(414.7)(535 – 108.83/2)
Mn = 368 x 106 kN.m. (nominal moment strength).

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