Solution - dm2_080341

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At t = 0.05891 second:
Pcr = (18.6/28.3)396.9
= 260.9 psi
Ps = (6.9/28.3)396.9
= 96.8 psi
Vu = (572.7 + 396.9)(2)28.3 +
2
(396.9 + 260.9)9.7
2
= 30,630 lb/in
Mu = 96.8(21.4)2 + (396.9 - 96.8)(21.4)2
2
3
+ 396.9[2(28.3)][2(28.3)/2 + 21.4]
+ (572.7 - 396.9)28.3[4(28.3)/3 + 21.4]
= 1,478,660 in-lb/in2
The peak shear occurs at t = 0.05891 second.
The corresponding bending moment at the critical section
for shear is computed as follows:
Mcr = 260.9(9.7)2 + (396.9 - 260.9)9.72
2
3
+ 396.9[2(28.3)][2(28.3)/2 + 9.7]
+ (572.7 - 396.9)28.3[4(28.3)/3 + 9.7]
= 1,106,179 in-lb/in
The peak shear and corresponding bending moment at the
critical section for shear are:
Vu = 30,630 lb/in
Mcr = 1,106,179 in-lb/in
The peak moment at the face of the support is:
Mu = 1,478,660 in-lb/in
(11) Determine the allowable shear stress for the concrete.
Since *ln/d < 5, the allowable shear is computed using
Equation (156a):
Vu
=
30,630 lb/in
Mcr
=
1,106,179 in-lb/in
d
=
90 inches
f'c
=
4,000 psi
pw
=
0.0025, assume minimum from Table 28
vc
=
0.85[3.5 - 2.5(1,106,179)/30,630(90)]
x [1.9(4,000)1/2 +
2,500(0.0025)(30,630)
x 90/1,106,179]
= 0.85(2.5)(135.7)
= 288.5 psi