Problem 2 This problem involves beam B3 from a structure with the plan view as s

Question

Problem 2 This problem involves beam B3 from a structure with the plan view as shown below. The cross- section of the beam is also given (with dimensions). The distributed dead and live loads along member B3 have already been calculated, and they are w-1350 plf and wLL-900 plf. The width of the webs of the girders (G1, G2, G3...) are all 16 a) Determine the effective flange width of beam B3. b) Determine the largest factored positive moment, Ma, on beam B3 (refer to section 11.8 in c) Determine the factored negative moment, Mu, on member B3 where it intersects with d) Determine the factored negative moment, M., on member B3 where it intersects with your textbook and Table 11.1). member G 1 . member G2. 20 11-0" Beam B3 29-0 G4 31-0

Explanation / Answer

The size of cross sectional beam = 20-5.5 = 15.5' d x12' w

The total load on beam B3= DL+LL = 1350+900 = 2250plf

The web width =16"

Since. it's a T-Beam, the lesser of    , bf ?= lo/6 +bw +df = 29/6 + 16" +15.5" =36" or 16" ; whichever is less.

Hence bf= 16"

b. The factored moment , governing equation is : - (as Xumax is greater than Df)

Mu = 0.36 fck bw xu, max (d - 0.42 xu, max ) + 0.45 fck (bf - bw) Df (d - Df /2)

or Mu = 0.36(xu, max /d){1 - 0.42( xu, max/d)} fck bw d2 + 0.45 fck(bf - bw) Df(d - Df)

Therefore Mu = 0.36*25*16"( 15.5- 0.42*10) + 0.45*25(16"-16")) *5.5( 20-5.5)

= 11.3 +0= 11.3

c) Mu = 2250 *23 = 51750 = 51.7Kn-m

d) Mu = 2250 * 23= 51750 =51.7Kn-m

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