For the beam shown in Fig.I, four conccntratcd live loads, each p1 = 15 kips (as
ID: 1844991 • Letter: F
Question
For the beam shown in Fig.I, four conccntratcd live loads, each p1 = 15 kips (as shown in Fig I) and a concentrated live load moments of Mt = 45 kip-ft acts as shown. The self wt. of the beam is not shown in the drawing. The maximum depth of the beam is restricted to h =18". USD Method for the design Design the beam for maximum positive and negative flexure (bending). Show the details of reinforcement placement. Given:f'c = 5,000 psi,fy = 60,000 psi. Note: The self wt. of the beam needs to be included. You may start with a size of 10" times 18" for the beam.Explanation / Answer
Solution: Given; three PL=15 Kip and two ML=45 Kip-ft
lets assume L/d = 1.5 due to heavy loads
Step 1: depth of beam = 18"
step 2: Effective span
width of support < L/12
Effective span = c/c distance of SSB = 8ft
Step 3: Loads
Dead load = self wt + superimposed loads
self wt = density* b* d
=25 * 0.8 * 1.5 = 30kip
Total DL = 30 Kip
total live load = 45 kip
Step 4: Design moments
At supports
Mu(-ve) = 1.5(WDL2/10 + WLL2/9)
= 768Kip-ft
+ve BM at centre of span
MU(+ve) = 1.5(WdL2/12 + WLL2/10)
= 456 Kip-ft
Mu,lim = Ru * bd2
= 700Kip
therefore Mu > Mu,lim
hence,it will be doubly reinforced
Step 6: Design of doubly reinforcement
(a) At supports
Ast,1 = Mu,lim/0.87 *fy *(d - 0.42 xu,max d)
= 335.92
Ast,2 = Mu - Mu,lim /0.87fy(d - d')
= 74.43
therefore total Ast = 410.35
therefore no. of 20 mm fi bars = 6 bars
now, Asc = Mu - Mu,lim/ fsc(d-d')
= 522.87
therefore no. of 20 mm fi bars = 2 bars
and at midspan = 6 bars