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