Finding the right beam

1. May 3, 2010

joemama69

1. The problem statement, all variables and given/known data
A simple beam of length L = 5m carries a uniform load of intensity q = 5.8kN/m and a concentratexd load 22.5kN. Assuming stress allow = 110MPa, calculate the required section modulus S. Then select a 200mm wide flange beam (W shape) from the table. recalculate S taking into account the weight of the beam. select a new 200mm beam if neccessary.

2. Relevant equations

3. The attempt at a solution

Of first I was trying to calculate the Mmax and i made a boo boo somwhere.

First I found the Moment about A

0 = 5.8(5)(2.5) + 22.5(3.5) - 5RB.... RB = 30.25kN

0 = 22.5(1.5) + 5.8(5)(2.5) - 5RA.... RA = 21.25kN

Then so determine the Maximum Moment I was preparing the Shear & moment graphs. I took x starting from the left where x = 0 @ A

0<x<3.5

Force sum in y = 0 = RA - qx - V.... V = 21.25 - 5.8x
Sum of M = 0 = RAx - qx2/2 - M... M = 21.25x - 2.9x2

3.5<x<5

Force in y = 0 = RA - qx - P -V.... V = -1.25 - 5.8x
Sum of M = 0 = RAx - qx2/2 - 22.5(x - 3.5) - M... M = 78.75 - 1.25x - 2.9x2

when I graph the V's, the lines do not intersect... did i do something wrong. Would it even matter as long as I did the M's correctly. Could someone just double check my work.

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2. May 3, 2010

PhanthomJay

Your equations look OK. Max moment occurs at point of zero shear. Draw a shear diagram. There is an abrupt change in shear at the applied concentrated load. If you plot your 2 shear equations, they won't intersect at a common point because of the concentrated load discontinuity.

3. May 3, 2010

joemama69

ok i found V = 0 @ x = 3.664

So pluged this into the second M equation and got Mmax = 35.7kN*m

S = M/$$\sigma$$ = 3.25X10^-4 what happened here

4. May 3, 2010

PhanthomJay

Your numbers are off a bit...point of zero shear ocurs at the concentratd load at x =3.5 m. M_max at that point is about 40 kN-m. Then when you calculate S, the result is in m^3. (i don't work in SI units, so i don't have a feel for the numbers)