HORIZONTAL SHEAR AT H/2

As for vertical shear, section in right half controls:

V_u = total factored shear force

= 1.3 ((S W + D L_{p r e c} + T o p + {D L}_{c o m p}) + 1.67 (L L)) = 1.3 ((43.26 k) + 9.57 k + 47.21 k + 17.10 k + 1.67 (80.70 k)) = 328.1 k

V_{n h} \geq \frac{V_{u}}{ϕ} = \frac{328.1 k}{0.90} = 364.6 k

Case (a): Contact surface is roughened.

V_{n h} = 80 b_{v} d = \frac{80 \times 42 \times 76.25}{1000} = 256.2 k < 364.6 k

Art. 9.20.4.3(a)

Case (b): Minimum ties provided, contact surface not roughened.

V_{n h} = 80 b_{v} d = \frac{80 \times 42 \times 76.25}{1000} = 256.2 k

Art. 9.20.4.3(b)

A_{v h \min} = \frac{50 b_{v} s}{f_{y}} = \frac{50 \times 42 \times 12}{60000} = 0.42 {i n}^{2} / f t

If 0.42 in²/ft. of steel is provided, capacity of unroughened surface is 256.2K. Therefore, additional steel must be provided to increase capacity. For each percent of tie reinforcement crossing the contact surface in excess of minimum, V_nh may be increased by:

\frac{(160 f_{y})}{40000} b_{v} d

Surf = (12)(42) = 504.0 in²/ft.

1% = (.01)(504.0) = 5.04 in²/ft.

Consequently, if 5.04 in² of steel are provided, additional capacity is:

= \frac{(\frac{60 \times 60000}{40000}) (42) (76.50)}{1000} = 771.1 K

Additional capacity needed is: = 364.1K -256.2K = 107.9K

Therefore, additional steel needed is:

\frac{107.9}{771.1} \times 5.04 = 0.70 {i n}^{2} / f t

A_{v h - s m} = 0.42 + 0.70 = 1.12 {i n}^{2} (s m o o t h c o n d i t i o n)

Case (c): Minimum ties provided, and contact surface roughened.

V_{n h} = 350 b_{v} d

Art. 9.20.4.3(c)

= \frac{(350) (42) (76.25)}{1000} = 1120.9 k < 364.6 k

If 0.42 in²/ft. of steel is provided, capacity of roughened surface is 1120.9K. Therefore, additional steel must be provided to increase capacity.

Additional capacity needed is = 364.6K -1120.9K = -756.3K

Additional capacity needed < 0, therefore no additional reinforcement is required

A_{v h - r g} = A_{v h - \min} = 0.42 {i n}^{2} / f t

S_{\max} \leq 4 b_{b w} o r 24 i n = 4 \times 6 o r 24 = 24 i n

Art. 9.20.4.5(a)

Therefore,

S_max = 24 in = maximum spacing of stirrups

V_{n h - r e q ` d} p s i = \frac{V_{u}}{ϕ b_{v} d} = \frac{327700}{0.90 \times 42 \times 76.25} = 113.7 p s i