# shear and moment diagram

A. determine Wo
B. Draw the shear and moment diagrams and specify the numerical values at all change of loading positions & at all points of zero shear.
C. determine the maximum shear and maximum moment.

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## First, it looks like this

UPDATEI did not see right away that the system is supported all throughout. I now appreciate the wisdom of this problem after I saw it. I will edit my response below to reflect the reaction that will be distributed from point

Ato pointD. For now, I have something I need to finish, so I will be back to this as soon as I can.OLD REPLY## Click here to expand or collapse this section

By inspection, I don't think the triangular reaction at

ABwill be equal to the triangular reaction atCD. The figure however suggest that they are equal because both are labeled withw. If you sum up forces in the vertical direction, all of the 50 kN get cancelled. What remains is the the spandrel load, the unsymmetrical placement of this load will make the triangular reactions to be unequal._{o}Another observation is that the origin for 6

x^{2}is unclear in the given. Isxmeasured fromAor fromB? Note that the load can be zero atBbut does not necessarily mean to be the vertex of the parabolic curve.With the errors in the given pointed out, I will attempt to answer this problem with the following premise:

wat_{o}ABto bew_{1}and atCDmay I call itw_{2}.whas a unit of kN/m.x^{2}is atBso that atC, the magnitude is 96 kN/mHere is my solution:

$\Sigma M_A = 0$

$\frac{2}{3} \cdot \frac{1}{2}(1)(w_1) + (5 + \frac{1}{3}) \cdot \frac{1}{2}(1)(w_2) = \frac{1}{3}(4)(96)\left[ 5 - \frac{1}{4}(4) \right]$

$\frac{1}{3}w_1 + \frac{8}{3}w_2 = 512$ ← Equation (1)

$\Sigma M_D = 0$

$(5 + \frac{1}{3}) \cdot \frac{1}{2}(1)(w_1) + \frac{2}{3} \cdot \frac{1}{2}(1)(w_2) = \frac{1}{3}(4)(96)\left[ 1 + \frac{1}{4}(4) \right]$

$\frac{8}{3}w_1 + \frac{1}{3}w_2 = 256$ ← Equation (2)

From Equations (1) and (2)

$w_1 = 73.14 \text{ kN/m}$

$w_2 = 182.86 \text{ kN/m}$