Question

Page of 12O The limit states design equation for flexure requires that фМ.-φhdek Mu . Initially, assume φ= 0.9 this will #4@18. VEF #4@18. HEF 16" End Zone (9)-# 8 VEF and #4 ties (typ. each end) 3/4" cover (lypical) FIGURE 8-32 Shear w all reinforcement detail. be checked later after et is determined) and then calcu-late the required k as follows: 1950(12)0.55 Mu hd (0.9)(8 in.)(76.8 in.) From Table A-10, we obtain p 0.0101 and e, > 0.005; therefore, 0.9, as initially assumed. The concen-trated vertical reinforcement required at each end zone of the shear wall is A, required = phd (0.01 0118 in)(76.8 in.) 6.21 in.2

I want to know the table that he uses to obtain rho it said table A-10 but I want to see that table because I don't know what table is that

Answer 1

M_u=1950 kips

b=8 in

d=76.8 in

M_u/( $\phi$ bd²)=1950*12*1000/(0.9*8*76.8²)=550 psi

To obtain the value of $\rho$ we need F_y and F_c values

(1) If Fy=50 ksi and Fc=3 ksi (use Table A.10)

(2)If Fy= 50 ksi and Fc=4ksi (use TableA.11)

(3)If Fy=60 ksi and Fc=3 ksi (use Table A.12)

(4) If Fy=60 ksi and Fc=4 ksi (use Table A.13)

From the tables shown above, the value for $\rho$ =0.101 is possible only for Table A.13, having K=550 psi

$\rho$ =0.0101

A_st=0.0101x8x76.8=6.2 in²

Here is the Table provided for reference: