Ultimate Moment Capacity of Beam

Problem 1. Determine allowable working moment of a singly reinforced beam having a section of the beam is 10 " * 10" and d=15 inch, use 3 - 20 mm dia bar. Given f_s = 20000 psi, f_c = 1350 psi and n = 9?
Exam Name: PGCL 2017


Solution:
Ultimate moment for concrete,
              M_c = (1/2) f_cjkbd²

    Here,   f_c = 1350 psi  
                 j = 1 - k/3
                k = n / (n + r)
                 r = f_s /f_c
                   = 20000/1350
                   = 14.81
           ∴  k = 9/ (9 + 14.81)
                  = 0.378
                j = 1 - (0.378/3)
                  = 0.874
               b = 10
               d = 5

     ∴     M_c = (1/2)*1350*0.874*0.378*10*15²
                  = 501752 lb
                  = 501.752 kip


          M_s =A_sf_sjd
                = 3* π/4*0.787²*20000*0.874*15
                = 383034 lb
                = 383 kip

∴ Working moment capacity 383 kip

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Problem 2.  A rectangular beam has width of 12 inch an effective depth to the centroid of the reinforcing steel of 17.5 in. It is reinforced with four no. 9 bar in one row. If f_y= 60 ksi and f_c'=4 ksi. What is the ultimate moment capacity of the beam? Also check what would be the value of φ ?
Exam Name: DESCO 2015


Solution:
Ultimate moment,
            M_u = φM_n = φA_sf_y(d-a/2)
  Here,  A_s = 4* π/4* 1.128²
                  = 4 inch²
               b = 12"
               d = 17.5"
           ∴ρ  = A_s  / bd
                  = 4 / (12*17.5)
                  = 0.019

 ρ_max = {(0.003 + f_y/E_s) / 0.008 } * {0.85 β₁ (f_c'/ f_y )}* {87/ (87 +  f_y)}
          =  0.018 <  ρ = 0.019

So the section is in transition region and  φ <0.9

Now,
         a = A_sf_y / 0.85f_c'b
            = {4*60/ (0.85*4*12)}
            = 5.88 in

        c = a/ β₁
           = (5.88/0.85)
           = 6.92
       ε_t = (d_t -  C)/ C * 0.003
           = {(17.5 - 6.92) / 6.92} * 0.003
           = 0.0046

       φ = 0.65 +{(ε_t - 0.002)* (250/3)}
          = 0.87

   ∴ M_u = φM_n = φA_sf_y(d-a/2)
             = 0.87 * 4 * 60 * {17.5 - (5.88/2)}
             = 3040.13 kip -in
             = 253.34 kip-ft

           

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