Strength of Materials Problems and Solutions Worked Step by Step

Strength of materials studies how solid members respond to loading: the internal stresses that develop, the deformation that results, and the conditions under which failure occurs. It covers axial loading, bending, shear, torsion, combined stresses, columns and pressure vessels, and it underpins the design of almost every load-bearing component.

Direct stress is the load divided by the cross-sectional area, giving N/mm² or MPa. Direct strain is the change in length divided by the original length, so it is dimensionless. Within the elastic range they are related by Hooke's law through Young's modulus.

Young's modulus is the ratio of direct stress to direct strain in the elastic range, so it measures stiffness rather than strength. For steel it is roughly 200 GPa, meaning 200,000 N/mm². A stiffer material deforms less under the same stress but is not necessarily stronger.

Numerically both are force per unit area and share units. Pressure acts normal to a surface and is usually externally applied by a fluid, while stress describes the internal distribution of force within a material and can be normal, shear or a combination of both.

It depends on the loading and supports. For a simply supported beam with a uniform load it is wL² divided by eight at mid-span; with a central point load it is PL divided by four; for a cantilever with a uniform load it is wL² divided by two at the fixed end.

Section modulus is the second moment of area divided by the distance to the extreme fibre. It combines both into one property, so maximum bending stress is simply the bending moment divided by it. Comparing section moduli ranks sections directly for bending capacity.

Because the second moment of area varies with the cube of depth but only linearly with width. Doubling the depth increases bending stiffness roughly eightfold, whereas doubling the width only doubles it. This is why joists are installed on edge rather than flat.

The second moment of area, I, governs bending and for a circle equals pi d to the fourth over sixty-four. The polar second moment, J, governs torsion and equals pi d to the fourth over thirty-two, so J is exactly twice I for a solid circular section.

Multiply the applied torque by the radius at which stress is wanted and divide by the polar second moment of area. The maximum occurs at the outer surface. Shear stress varies linearly from zero at the axis, which is what makes hollow shafts efficient.

Because torsional shear stress is near zero close to the axis, so the central material carries very little load while contributing its full weight. Removing it reduces mass substantially with only a modest loss of torsional capacity, improving strength for a given weight.

Principal stresses are the maximum and minimum normal stresses at a point, acting on the planes where shear stress is zero. They are found from the average normal stress plus or minus the radius of Mohr's circle, and they govern most failure criteria.

Mohr's circle represents the state of stress at a point graphically, so principal stresses, maximum shear stress and plane orientations can be read directly. The circle's centre is the average normal stress and its radius is the maximum in-plane shear stress.

Add them. The sum of the principal stresses must equal the sum of the original normal stresses, because the sum of normal stresses is invariant under rotation of axes. If the two sums disagree, there is an arithmetic error to find.

In a thin-walled cylinder under internal pressure, hoop stress acts circumferentially and equals pressure times diameter over twice the thickness. Longitudinal stress acts along the axis and is half that value, which is why pressurised pipes split along their length.

Yes. A one-to-one tutor can work through axial, bending, torsion and combined-stress problems with you, check your sign conventions and units, and show you the self-checks that catch errors. You complete your own assessed work in line with your course rules.