Determine the design moments for a two span beam that is continuous over three s
ID: 1713007 • Letter: D
Question
Determine the design moments for a two span beam that is continuous over three supports. (The ends are not intergral with the column supports and the column stiffness is less than 8 in all spans.) Determiine the critical design positive and negative moments for each span. Assume that the structure carries a uniformly distributed load of 475 lb/ft and that each span is 24 ft. Then determine the critical design shear force.
Answers: M+ end spans = 24,873 lb-ft, M- at interior support = 30,400 lb-ft, M- at exterior supports = 0 lb-ft; Vmax at interior = 6,555 lb
Explanation / Answer
For a two-span, continuous beam with constant moment of inertia throughout,
A). approximate critical design positive moment (where discontinuous end is unrestrained) is given by
,
Where l = length of the span, &
W = load per unit length.
Therefore, M = 475*24*24/9 = 30400 lb-ft
B). similarly, approximate critical design negative moment at exterior face of first interior support is given by,
Where l = length of the span, &
W = load per unit length.
Therefore, M = 475*24*24/11 = 24873 lb-ft
C). at the exterior supports, there are no moments.
Hence M(-) at the exterior supports = 0 lb-ft
For a two-span, continuous beam with constant moment of inertia throughout,
A). approximate critical design positive moment (where discontinuous end is unrestrained) is given by
,
Where l = length of the span, &
W = load per unit length.
Therefore, M = 475*24*24/9 = 30400 lb-ft
B). similarly, approximate critical design negative moment at exterior face of first interior support is given by,
Where l = length of the span, &
W = load per unit length.
Therefore, M = 475*24*24/11 = 24873 lb-ft
C). at the exterior supports, there are no moments.
Hence M(-) at the exterior supports = 0 lb-ft
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