Problem 2 A beam supports a distributed load as shown in the

Problem 2

A beam supports a distributed load as shown in the figure. Please do the following:

(a) Develop an expression (in symbolic form – do not substitute values at this time) for the distributed load along the length of the beam. Your expression should be general enough that it could apply to any linearly distributed load on the beam.

(b) Develop expressions, in symbolic form, for the equivalent concentrated load. Why multiple expressions (i.e., equations)? Well, in order for a load to be equivalent to the distributed load, don\'t we need to know more than just the magnitude? Do your equations make sense? Comment on why/why not? Provide insight.

(c) Develop expressions, in symbolic form, for the reaction force and reaction moment at the support (base of the cantilever beam). Note that you can find these expressions in the appendix of any undergraduate mechanics of materials textbook – but I am asking you to derive them in this problem. Show all of your work! But feel free to check your work against the published equations.

(d) Using the conditions in the figure, evaluate your expressions from (c) for the reaction force and moment at the base of the beam. Note: Although no variables names are provided in the figure, go ahead and create your own and use these in your analysis.

Solution

Let 600N/m be considered as b and 240N/m be considered as a.

a) Let the linearly distributed load be divided into two parts one uniformly distributed load a and other linearly varying load from 0 to b-a.

For the linearly varying part the value of load at a distance x from the free end can be calculated using similar triangles:

4.2/b-a = x/w_x

Hence w_x = (b-a)x/4.2

Hence the symbolic expression for distributed load at any distance x from the free end is a+(b-a)x/4.2

b) For the uniformly distributed load part the resultant concentrated load for a section of length x from free end acts at a distance x/2 m from the free end and its magnitude is ax.

For the linearly varying load part the resultant concentrated load for a section of length x from the free end acts at a distance of 2x/3 from the free end and its magnitude is (b-a)x/2.

Hence symbolic expression for concentrated load at a distance x is difficult to obtain rather it is obtained for a section of length x from free end.

In this case the symbolic expression of load using weighted average method is (ax*(x/2)+(b-a)x/2*2x/3)/x

Hence it is clear that for determining the concentrated the information about magnitude is not sufficient. We also require information about the type of load distribution which is important for determining the point of application of force.

c) Using the condition of vertical force equilibrium

R_A= a*4.2+(b-a)*4.2/2 = 2.1(a+b)

M_A = a*4.2(2+2.1)+(b-a)/2*4.2(2+4.2/3) = 10.08 a+ 7.14b

d) Hence subsituting the values of a and b

R_A = 2.1(240+600) = 1764 N

M_A = 10.08*200+7.14*600 = 6703.2 Nm

which is equal to the values obtained by considering the loads in the diagram above. Hence it verifies symbolic expressions.