In: Civil Engineering
A solid grouted 8 in masonry CMU beam is reinforced with two #4 Grade 60 bars andhas an effective depth of 32.5 in. Determine the allowable moment in accordance withACI 530. Use f’m = 2500 psi and assume medium weight units (140 pcf). What is theallowable load if the effective span is 10.67 ft?
Ans) As per ACI 530 , the allowable moment (M) in CMU beams is given by,
M = (Asfy + P ) ( d - a/2)
where , As = area of steel
fy = yield strength of bar ( for grade 60 rebars fy = 60000 psi)
P = medium weight unit x volume
d = effective depth
a = (P + Asfy ) / (0.8 f'm b)
Area of steel = 2 x /4 x D2
For #4 grade 60 bar, Diameter = 0.5 in
Therefore , As = 2 x 0.785 x 0.5 x0.5
= 0.3925 in2
a = [(140 / 123) x 34.5 x 10.67 x12 x 8 ) + (0.3925 x 60000)] / (0.8x 2500 x 8)
On solving, a = 1.65
Putting all values to calculate allowable moment,
M = [(0.3925 x 60000) + (140/123)] ( 32.5 - 1.650/2)
M = 745922.70 lb-in
To determine allowable load(F), we know
F = (M y /I) x bd
where, I = moment of Inertia
y = distance of neutral axis from base of beam ( y = d/2)
We know,
I = b d3 /12
= 8 x 34.53 / 12 ( 2 in added in depth for cover)
= 27375.75 in4
Hence,
F = [(745922.70 x 34.5/2) / 27375.75] x 32.5 x 8
= 122205.35 lb