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Chapter 3

Q. 3.8

Our company designs and manufactures pre-engineered steel buildings. The structural system uses tapered steel members to create a rigid frame with a pitched roof. The members become thinner near the top, so the frame behaves like athree-hinge arch.

To design the connection between the vertical and diagonal members, we need to know the moment transferred at that point. One of the load cases we need to check is heavy snow on only one side of the roof.

E3.65
E3.666

Step-by-Step

Verified Solution

Analysis:
FBD of AC:

+\circlearrowright ΣM_A = 0 = 0.5  klf  (45  ft)  (45  ft/2) + C_x  (40  ft) + C_y  (45  ft)

C_x  (40  ft) = -C_y  (45  ft)  –  506  k  •  ft

C_x = -1.125C_y  –  12.65  k

FBD of CE:

+\circlearrowright ΣM_E = 0 = -C_x  (40  ft) + C_y  (45  ft)

C_x  (40  ft) = C_y  (45  ft)

C_x = 1.125C_y

Combine the C_x expression with the result from AC:

1.125C_y = -1.125C_y  –  12.65  k

2.25C_y = -12.65  k

C_y = -5.62  k

Substitute C_y into the expression for C_x:

C_x = 1.125  (-5.62  k)  ;      C_x = -6.32  k

FBD of BC:

+\circlearrowright \sum{M_B} = 0 = M_{BC} + 0.5  klf  (45  ft)(45  ft/2) + (-6.32  k)(20  ft) + (-5.62) (45  ft)

M_{BC} = -127.0k  •  ft

FBD of CD:

+\circlearrowright \sum{M_D} = 0 = (-6.32  k)(20  ft)  –  (-5.62)(45  ft) + M_{DC}

M_{DC} = -126.5  k  •  ft

Summary:        M_{BC} = -127.0  k  •  ft

        M_{DC} = -126.5  k  •  ft

E3.67
E3.68
E3.69
E3.70