Homework SourseUsing plastic theory and A992 steel determine the value of P
Solution
Given:
Beam Section : W 24 x 94
Section modulus Zx : 172 in2
Yield strenght of steel fy : 50 ksi
To Find : Uniform Load wu
Solution:
The nominal plastic moment of the beam is calculated first
Mn = fy Z = 50 x 172 = 8600 ft-k
The virtual -work expression written separatly for each span of the continous beam
The collapse machanisms for three span are drawn
Locate the plastic hinge from left span
X = 0.414L = 0.414 x 24 = 9.936 ft
From the triangle of the right span:
tan \\Theta \\Theta = \\delta2/14 \\delta2 = 14 \\Theta
\\Theta = \\delta1/12 \\delta1 = 12 \\Theta
From the triangle of the left span
24-x = 24-9.936 = 14.064 ft
Also
tan \\Theta \\Theta = \\delta3/14.06 \\delta3 = 14.06 \\Theta
Therefore
tan\\alpha = \\alpha = \\delta3 / 9.936 = 14.06 \\Theta /9.936 = 1.415 \\Theta
Virtual -work applied to right span (span 1)
External work = Internal work
Wext = Wint
(wn L1 ) x ( \\delta1 )avg = Mn (\\Theta+\\Theta+2\\Theta)
(wn x 24 ) ( 1/2 x 12\\Theta) = Mn (4\\Theta)
wn1 = 2 x (4x Mn) /(24 x 12) = 0.0278 Mn = 0.0278 x8600 = 238.9 k/ft
Virtual -work applied to right span (span 2)
External work = Internal work
Wext = Wint
(wn L2 ) x ( \\delta2 )avg = Mn (\\Theta+\\Theta+2\\Theta)
(wn x 28 ) ( 1/2 x 14\\Theta) = Mn (4\\Theta)
wn2 = 2 x (4x Mn) /(28 x 14) = 0.020 Mn = 0.020 x8600 = 175.5k/ft
Virtual -work applied to left span
Wext = Wint
(wn L3) x ( \\delta3 )avg = Mn (\\alpha+\\Theta+\\Theta)
(wn x 24 ) ( 1/2 x 14.06\\Theta) = Mn (1.415\\Theta+\\Theta+\\Theta)
wn3 = 2 x (3.415x Mn) /(24 x 14.06) = 0.020 Mn = 0.020 x8600 = 174.1 k/ft
Result :
The uniform load wu = 174.1 k/ft
