Problem #4: Given a singly reinforced beam with fy-60 ksi,fc-5,500 psi, b= 16\",

Question

Problem #4: Given a singly reinforced beam with fy-60 ksi,fc-5,500 psi, b= 16", d-25". If the ultimate moment Mu 680 ft-kips, compute the required area of tensile stee, "As-required". Choose an optimum number of bars (just use one bar size not a combination of bar sizes) using the #3 thru #10 range. Also use more than 2 bars but no more than 14 bars. Compute the Mn of the final section (and of course it should be greater than or equal to the required Mu.) Note that the exact value of is unknown until the actual As is determined. There are several solutions, do they really fit?

Explanation / Answer

Solution:-

Given

b= 16 inch

d =25 inch

Mu = 680 feet –kips

We know that

Xm/d =0.48

Xm = 0.48*25 = 12 inch

Ast = Mu/(0.87*fy*(d – 0.42Xm)

       = 680*12*103/(0.87*60000*(25 – 0.42*12)

   = 7.83 inch2

#9 bars are used

Diameter of bars is 1.128 inch

Total number of bars = 7.83/(/4*(1.128)2)

= 8

Total 8 bars are used.

Computation of Mn

Mn = Astfy(d – a/2)

Calculation of a :-

0.85fck *a*b =Ast*fy

a = 7.83*60000/(0.85*5500*16)

   a = 6.28 inch

Mn = 0.9*7.83*60000*(25 – 6.28/2)

        = 9.242845*106 lb-inch

        = 770.23 feet –kips    Answer

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