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Calculate the maximum B.M the symmetrical section can carry Plyweb with timber flanges (50x100 c.s.a) The maximum stress = 1.0N/mm^2
The structure is essentially an I beam with dimensions as follows.
The central beam has a width of 25 and height of 500.
The four flanges which constitute the I shape have each a width of 50 and height of 100 and are flush with the top and bottom of the central beam.
Since no units are given I have assumed all dimensions are in mm.
B.M = W.L / 4 The Answer = 4.3kNm
Total force acting upon the beam is the udl of 1.0N/mm2.
Total width is 125mm therefore w = 125x1 = 125 N.
W.L / 4 = 125^2 / 4 = 3906.25 Nm which is clearly wrong with respect to the answer given.
Working backwards. 4300 N = W.L / 4 gives (4300 x 4) / L (where L = 125) gives a value of W = 137.6
Any help would be appreciated as I am majorly confused over such a simple problem.
The structure is essentially an I beam with dimensions as follows.
The central beam has a width of 25 and height of 500.
The four flanges which constitute the I shape have each a width of 50 and height of 100 and are flush with the top and bottom of the central beam.
Since no units are given I have assumed all dimensions are in mm.
B.M = W.L / 4 The Answer = 4.3kNm
Total force acting upon the beam is the udl of 1.0N/mm2.
Total width is 125mm therefore w = 125x1 = 125 N.
W.L / 4 = 125^2 / 4 = 3906.25 Nm which is clearly wrong with respect to the answer given.
Working backwards. 4300 N = W.L / 4 gives (4300 x 4) / L (where L = 125) gives a value of W = 137.6
Any help would be appreciated as I am majorly confused over such a simple problem.