Design of a Beam Subject to Torsion -Cont. - dm2_080136

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First, check that the requirement of Equation (55) is met:
2At = 2 x .08
= .16 in2/ft
50bs
50 x 18 x 12
=
fdy
66,000
= .16 in2/ft < /= .16 in2/ft
O.K.
Therefore,
400 x 18 x 12
97.1
A*l =
() - 0.16
60,000
97.1 + 202.7
14.5  +26.0
x   =  1.03 in2
12
= 1.03 in2 > 0.54 in2
O.K.
(9) Distribute A*l, As, and As as follows (see Figure
39):
Distribute A*l equally between four corners of the beam
and one on each face of depth, a total of six locations
to satisfy maximum spacing of 12 inches.
A*l/6 = 1.03/6
= 0.17 in2
Vertical Face:
One No. 4 bar = 0.20 in2 > 0.17
O.K.
Horizontal Face at Top:
Support = 2.20 (bending) + 2 x 0.17 (torsion)
=  2.54 in2
Two No 7 at corners + three No. 6
= 2.52 in2  O.K.
Midspan = 1.64 (rebound)
Two No. 7 at corners + one No. 6
= 1.64 in2  O.K.
Horizontal Face at Bottom:
Support = Greater of rebound (1.64 in2)
or torsion (2 x 0.17)
Two No. 7 at corners + one No. 6
= 1.64 in2  O.K.
Midspan = 2.20 (bending)
Two No. 7 at corners + one No. 6 + two No. 5
= 2.26 in2 > 2.20  O.K.
2.08-119