Quantcast Solution -Cont. - dm2_080129

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2 x 2.10
tan [theta] = = 0.0175
240
[theta] = 1.0[deg.] < /= 1[deg.]
O.K.
(13) Direct shear from Table 9 is:
ruL
Vs =
2
1,065.85 x 240
=
2
= 127,902 lb
Section capacity in direct shear from Equation (46):
Vd = 0.18 f'dc bd
= 0.18 x 4,400 x 18 x 27.125
= 386,694 lbs > Vs = 127,902
O.K.
(14) Diagonal tension stress from Equation (43):
Vu
vu =
< /= 10[phi] (f'c)1/2
bd
Total shear d distance from the face of support:
Vu = (L/2 - d) ru
240
= (  - 27.125) 1,065.85
2
= 98,990 lb
98,990
vu =
18 x 27.125
= 202.7 psi
10 [phi] (f'c)1/2 = 10 x 0.85 x (4,000)1/2
= 537.6 psi > 202.7 psi  O.K.
(15) Unreinforced web shear capacity using Equation (45) is:
vc = [phi] [1.9 (f'c)1/2 + 2,500 p]
= 0.85 [1.9 (4,000)1/2 + 2,500 x 0.0045]
= 111.7 psi
2.28 [phi] (f'c)1/2 = 2.28 x 0.85 x (4,000)1/2
= 122.6 psi >/= 111.7 psi  O.K.
(16) Area of web reinforcing from Equation (45):
Av = [(vu - vc) x b x ss]/[phi] fy
2.08-113





 


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