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Strength of Materials Problem Solving Quick Reference Guide

R. Strong ENGR 213 Portland Community College Spring 2011


Shaft Torque vs. Twist Shafts are often used as “straight springs” such as for torsion bars in automobiles (bottom right figure). When used as the spring element in a suspension system, they are made from heat-treated high-carbon steel and are sized according to the expected load capacity and desired wheel travel. The angle of twist is a function of the length (l) of the torsion bar spring, the applied torque (T) (bottom left figure), the material’s modulus of rigidity (G), and its polar moment of inertia (J). A round cross-section is highly recommended since it is the most efficient from a weight perspective and minimizes stress concentrations. To calculate the twist angle of a round shaft, use the following formula: T = Torque (Force x moment arm) l = length of shaft ! = (! r4)/2 (“r” is radius, or one-half the shaft diameter G = Modulus of Rigidity (table look-up value for mat’l) Note that this formula will give an angle in radians. To convert to “degrees”, multiply your result by 57.3. To calculate the wheel travel, multiply Φ (in radians) by the arm length extending from the spring out to the wheel. Care must be taken to ensure that the allowable limit of sheer (τmax) is not exceeded once a preliminary design is chosen using: τmax = Tr/J

Shear Element Visualization

Torsion Bar Front Suspension


Twist Calculation Problem:


Beam Deflection Beams comprise a substantial percentage of mechanical and structural design. Even if loads do not result in internal stresses that exceed the allowable limits of the material, a design can still be considered a failure if there is excessive deflection. Consider a simply supported bridge one might build that spans a creek running across a property’s main access. If the bridge were to “sag” more than six inches, the occupants in a crossing vehicle may feel discomfort and sense an immanent collapse even if there is plenty of strength margin. Determining the maximum deflection begins with selecting the appropriate “end conditions” of the beam. Fixed ends, for example, will exhibit less than 1/3 the center deflection of a design whose ends are free to pivot. For a simply supported bridge constructed from two beams spanning a gap (covered by cross-wise planks to create the driving surface), use the following formula: Deflection: Ymax = F * l3 / 3EI I = moment area of inertia. For rectangular beams I = 1/12 base x height^3. From this formula, it is obvious a beam will be much stronger when with the longer of the two dimensions oriented vertically. For round beams such as the log beams shown in the photo below, I = ! d4/64. Commercially formed beams such as U channels, W beams, and pipe have “I” values listed in any mechanical reference book. Remember to factor in the “total I” value by adding linearly the number of beams. F = Weight of the load (presume a center load for worst case) E = Modulus of Elasticity (steel ~ 30 x 106 psi, alum ~ 10 x 106 psi, wood such as Douglas fir ~ 2 x 106 psi)


Beam Deflection Problem:


Bending Stress- Flexure Formula

Shear & Moment Diagrams Beam Sizing Area Moment of Inertia Column Analysis Stress Concentrations Thin-walled Pressure Vessels Plane Stress Analysis- Mohr’s Circle Bending Stress- Flexure Formula Shear & Moment Diagrams



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