Problem 1

a )

The strain curvature relationship is....

8 Pts

By Hooke's law corresponding stress is...

Consider the figure on the left.
The total moment about the z-axis can be written as...

Substitute for the stress...

is the moment of inertia...

Hence ...

b )

From equilibrium the total axial forces in x direction should vanish. Hence...

2 Pts

which gives...

If stress is substituted...

If the material is homogenous and since we are considering pure bending E and k are constant. Then...

Hence...

If the axis are chosen such that above relationship hold then that axis passes thorough the centroid of the axis. And this axis where the axial stress vanishes is called the neutral axis.

c )

i )

Conceder the free body diagram on the left. From symmetry...

Then the moment between the supports is...

8 Pts

ii )

From the relations derived in part ( a )...

4 Pts

Since the moment between the supports is constant the curvature is constant between the supports as well...

iii )

Consider the deformed shape shown on the left.
The deflection in the middle can be expressed as...

... (1)

For small angles cos(q) can be approximated as...

... (2)

Since k is constant, then the deformed shape is perfect circle, hence using the small deformation approximation...

... (3)

Using eqns (1) , (2) and (3) we can get...

8 Pts

Then using...

Problem 2)

Consider the free body diagram on the rigth

2 Pts

The shear force diagram can easily be since all the loading and the reactions on the beam are known.

The point where the maximum moment occurs is the point where shear force vanishes. The location of that point can be obtained by using trianglar similarity.

giving ...

1 Pt

Than the moment diagram can be plotted using the shear force diagram.

2 Pts

The moment of inertia has to be calculated. This requires determination of the centroid of the section. Since the section is not doubly symmetric then the centroid will not be at h/2.

5 Pts

The moment of inertia for the section can be found using the parallel axis theorem.

5 Pts

Then the maximum stresses are....

5 Pts

kPa

5 Pts

Pa

If the section is rotated as shown in the figure on the left the sectional modulus would increase since it is proportional to the square of the height.

5 Pts

and S a

By rotating the section as such the heigth of the section will be increased which will reduce the maximum bending stress.