Determine the following Loads developed in each column suppo

Determine the following: Loads developed in each column support. (1A, 2A, 3A, 1B, 2B, 3B, 1C, 2C, 3C) Live Load (LL) = 60 psf Dead Load (DL) = 8 psf Column wt. 100 lbs Beam wt. = 10 lbs/ ft.

Solution

Tributary area of column 1A = (6/2)*(16/2)=24 sq ft

Tributary area of column 1B = (8/2 + 6/2)*(16/2)=56 sq ft

Tributary area of column 1C = (8/2)*(16/2)=32 sq ft

Tributary area of column 2A = (16/2 + 12/2)*(6/2)=42 sq ft

Tributary area of column 2B = (16/2 + 12/2)*(6/2 + 8/2)=98 sq ft

Tributary area of column 2C = (16/2 + 12/2)*( 8/2)=56 sq ft

Tributary area of column 3A = (12/2)*(6/2 )=18 sq ft

Tributary area of column 3B = (12/2)*(6/2 + 8/2)=42 sq ft

Tributary area of column 3C = ( 12/2)*( 8/2)=24 sq ft

Dead load intensity=8 psf

Live load intensity = 60 psf

Load on columns from unifrom distributed loads:

1A : (60+8)*24=1.632 kips

1B: (60+8)*56=3.808 kips

1C: (60+8)*32=2.176 kips

2A: (60+8)*42=2.856kips

2B: (60+8)*98=6.664 kips

2C: (60+8)*56=3.808 kips

3A: (60+8)*18=1.224 kips

3B: (60+8)*42=2.856 kips

3C:(60+8)*24=1.632 kips