Mechanics of Materials Flashcards

A solid circular steel shaft with a diameter of 50 mm is subjected to a pure torque of 2.5 kNm. What is the maximum shear stress in the shaft?

Answer: 101.9 MPa

The maximum shear stress in a solid circular shaft is calculated using the torsion formula: τ_max = (T * r) / J. First, calculate the polar moment of inertia, J = (π/2) * r^4. Given diameter d = 50 mm, the radius r = 25 mm = 0.025 m. So, J = (π/2) * (0.025 m)^4 = 6.136 x 10^-7 m^4. The torque T = 2.5 kNm = 2500 Nm. Now, calculate τ_max = (2500 Nm * 0.025 m) / (6.136 x 10^-7 m^4) = 101,861,147 Pa ≈ 101.9 MPa.

A 2-meter-long aluminum rod (E = 70 GPa) has a circular cross-section with a diameter of 20 mm. If a tensile load of 30 kN is applied axially, what is the resulting elongation of the rod?

Answer: 2.73 mm

The elongation (δ) of a rod under axial load is given by the formula δ = PL / AE. First, calculate the cross-sectional area A = π * r^2 = π * (0.01 m)^2 = 3.1416 x 10^-4 m^2. The load P = 30 kN = 30,000 N. The length L = 2 m. The modulus of elasticity E = 70 GPa = 70 x 10^9 Pa. Plugging these values in: δ = (30,000 N * 2 m) / (3.1416 x 10^-4 m^2 * 70 x 10^9 Pa) = 0.002728 m, which is equal to 2.73 mm.

A cantilever beam of length L has a point load P applied at its free end, resulting in a maximum deflection δ. If the length of the beam is doubled to 2L while the load P and cross-sectional properties (E, I) remain constant, what is the new maximum deflection?

Answer: 8δ

The formula for the maximum deflection of a cantilever beam with a point load at the free end is δ_max = PL^3 / (3EI). Deflection is directly proportional to the cube of the length (L^3). If the original length is L, the deflection is δ_1 = PL^3 / (3EI). If the new length is 2L, the new deflection is δ_2 = P(2L)^3 / (3EI) = P(8L^3) / (3EI) = 8 * (PL^3 / (3EI)) = 8δ_1. Therefore, the deflection increases by a factor of 8.

Which of the following statements best describes the shear and moment diagrams for a simply supported beam subjected to a single concentrated counter-clockwise moment at its mid-span?

Answer: The shear diagram is a horizontal line (constant non-zero value), and the moment diagram consists of two sloped lines with an abrupt vertical jump at mid-span.

An applied concentrated moment does not create shear itself, but it requires vertical reactions at the supports to maintain equilibrium, resulting in a constant, non-zero shear force across the entire beam. The moment diagram (the integral of the shear diagram) will therefore be linear (sloped). At the point where the concentrated moment is applied, there is an abrupt vertical jump or drop in the bending moment diagram equal to the magnitude of the applied moment.

A point on a structural element is subjected to a state of plane stress with the following components: σ_x = 80 MPa, σ_y = -40 MPa, and τ_xy = 30 MPa. What is the maximum principal stress (σ_1) at this point?

Answer: 87.1 MPa

The principal stresses can be found using the formula: σ_1,2 = ( (σ_x + σ_y)/2 ) ± sqrt( ( (σ_x - σ_y)/2 )^2 + τ_xy^2 ). The average normal stress is (80 + (-40))/2 = 20 MPa. The term under the square root (which is the radius of Mohr's Circle) is sqrt( ( (80 - (-40))/2 )^2 + 30^2 ) = sqrt(60^2 + 30^2) = sqrt(3600 + 900) = sqrt(4500) = 67.1 MPa. The maximum principal stress σ_1 is the average stress plus the radius: σ_1 = 20 MPa + 67.1 MPa = 87.1 MPa.

A 5-meter-long steel column (E = 200 GPa) with a moment of inertia of 2.53 x 10^6 mm^4 is pinned at both ends. What is the approximate critical buckling load for this column?

Answer: 200 kN

The critical buckling load is calculated using Euler's formula: P_cr = (π^2 * E * I) / (K * L)^2. For a column with pinned ends, the effective length factor K = 1.0. First, ensure consistent units: I = 2.53 x 10^6 mm^4 = 2.53 x 10^-6 m^4. E = 200 GPa = 200 x 10^9 N/m^2. L = 5 m. Now, substitute the values: P_cr = (π^2 * (200 x 10^9 N/m^2) * (2.53 x 10^-6 m^4)) / (1.0 * 5 m)^2 = 4,995,035 / 25 = 199,801 N ≈ 200 kN.