Respuesta :
force
F1 + F2 = 190 kN
8w2 + 8 ( w1 –w2)/2 = 190
W1 + w2 = 47.5 kN >>> equation 1
Torque
M=0 = 60(1) + 80(3.5) + 50 (7) – F1(4) – F2(2.667)
0 = 690 – F1 (4) – F2(2.667)
4F1 + 2.667F = 690
4( 8w2) + 2.667( 4w1 – 4w2) = 690
32w2 + 10.668w1 – 10.668w2 = 690
-2w2 - w1 = - 64.7 >>> equation 2
Solving equation 1 and 2
W2 = 17.2 kN
W1 = 30.3 kN
The intensities that w1 and w2 of this distribution needed to support the column loadings is 17.2 kN and 30.3 kN respectively
Force
F1 + F2 = 190 kN
8w2 + 8 ( w1 –w2)/2 = 190
W1 + w2 = 47.5 kN.............. Equation 1
Torque
M=0 = 60(1) + 80(3.5) + 50 (7) – F1(4) – F2(2.667)
0 = 690 – F1 (4) – F2(2.667)
4F1 + 2.667F = 690
4(8w2) + 2.667( 4w1 – 4w2) = 690
32w2 + 10.668w1 – 10.668w2 = 690
-2w2 - w1 = - 64.7.........Equation 2
Solving for Equation 1 and 2.
W2 = 17.2 kN
W1 = 30.3 kN
In conclusion, the intensities that w1 and w2 of this distribution needed to support the column loadings is 17.2 kN and 30.3 kN respectively
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