Question

Problem 2 (65%) (see table on page 2 for values of Wu(AB) and W (AC) Two simply supported beams meet at the corner 16 foot tall beam-column "A" shown. Beam AB carries supports a uniformly distributed factored design load W AB)=?? and is 24 feet long. Beam AC carries a uniformly distributed load, W.AC=?? and is 16 feet long. Compute the following if the beams and the beam-column are laterally supported at the ends only in this braced frame. a) Maximum shear force in beam AB b) Maximum bending moment in beam AB. c) Design the most economical W-shape for beam AB. d) Maximum shear force in beam AC e) Maximum bending moment in beam AC. f) Design the most economical W-shape for beam AC B) Total axial force, P, acting on the beam-column "A". Beam- h) Strong axis moment on the beam-column, M. - VAc xe, Column A i) Weak axis moment on the beam-column, My = VAB X 2 Double Angle 1) Use the AISC manual to design a simple shear connection suitable Bolted Shear for beam AB to the beam-column. Connection k) Use the AISC manual to design a simple shear connection suitable Beam for beam AC to the beam-column. AB Design the most economical W-shape to support the reactions from beam AB and AC Interaction equation check must be between 0.95 and 1.00 Beam АС -Nu(AB) (kips/ft) = 35 Nu(AC) (kips/ft) = 7

Answer 1

Solution:-

Given that

Wu(AB) (kips/ft) = 35

Wu(AC) (kips/ft) = 7

Beams are simply supported

$V_{max}=\frac{wL}{2}$ and $M_{max}=\frac{wL^2}{8}$

$V_{max}$ acts at ends & $M_{max}$ at center of beam.

a)

Wu(AB) = 35 k/ft

$V_{AB}=\frac{wL}{2}$

$=\frac{35\times 24}{2}$

= 420 kips

b)

$M_{AB}=\frac{W_u L^2}{8}$

$M_{AB}=\frac{35\times 24^2}{8}$

= 2520 kip . inch

c)

taking steel with yield stress,

(lateral support st ends only)

section modulus required

  $S_x=\frac{C_bM}{\phi F_y}$

$=\frac{2520 \ kip . inch\times 1.14}{0.9\times 36 \ kip/in^2}$

d)

$V_{AC}=\frac{W_u(AC)L_{AC}}{2}$

$=\frac{7\times 16}{2}= 56 \ kips$

e)

$M_{AC}=\frac{wL^2}{8}$

$M_{AC}=\frac{7\times 16^2}{8}= 224 \ k . ft = 2688\ k.inch$

f)

$S_x$ required = $\frac{C_bM}{\phi F_y}$

  $=\frac{2688\times 1.14}{0.9\times 36}$

= 94.57 in3

g)

Total axial force in column