1. ## Engineering statics

Calculate the internal shear (V), axial (N), and bending moment (M) in the beam at its mid-length.

Now heres what av figured out so far. Since its a fixed support at both ends there will be a vertical (V) and horizontal (V) reaction and also a bending moment (M). I got stuck from that point as when you take the moment about one point of the beam you have two unknowns (the Ms). Unless, since the beam is symmetrical, the vertical reactions are distributed equally between the two points? But still you would have the two unknown Ms. Any advice?

The distance (in mm) that the left hand end of the beam moves down under the given loading.

Dont understand how to do this question at all anyone able to help?

Cheers for the help

2. This is a pretty simple structural engineering problem. The problem is static determinate.

Note the left support of the beam is a pin and the right is a roller, which means there is no axial force in the beam (i.e., $N=0$)and there is no moment at both of its ends. There is only shear force at the ends.

By symmetry, there is no shear force at the middle of the beam ( $V=0$).

Hence, there is only moment at the middle of the beam.

Denote the peak of each triangular force by $q$, each concentrated force by $Q$, the total span of the beam by $L$, respectively, then the restraint force at either support is
$R=\frac{1}{4}qL+Q$

The moment at the middle of the beam is then
$M=R\times\frac{L}{2}-\frac{1}{4}qL\times\frac{L}{6}-Q\times\frac{1}{4}L=\frac{1}{12}qL^2+\frac{1}{4}QL$

Luobo