15 questions. Tap an option to check it instantly, then read the solution.
Practice SSC JE and other civil engineering exam questions on bending and shear stresses in beams: the bending equation M/I = σ/y = E/R and its assumptions, section modulus and moment of inertia of common sections, the strongest beam from a log, square versus circular sections, beams of uniform strength, shear stress distribution (rectangle, circle, I, T, angle, triangle, diamond), combined direct and bending stress, and a few related questions the book places in this chapter. Tap an option to check it, then open the full explanation for each question.
Question 1Medium
A short column carries an eccentric load. Which stress distribution at the base AB represents the case where the intensity of direct stress is less than the intensity of bending stress? (σmin = minimum resultant stress, σmax = maximum resultant stress)
Answer: C. Two triangles crossing AB: σmin tensile on one side, σmax compressive on the other
σmin = σ0 − σb. If σ0 < σb, σmin is negative (tension), so the diagram crosses the base line: tension on one side, compression on the other.
Question 2Medium
A unit rotation is to be caused at end A of a beam AB of length 5 m whose far end B is hinged. The moment that has to be applied at A is (EI = flexural rigidity):
Stiffness with far end hinged = 3EI/L = 3EI/5 = 0.6 EI.
Question 3Easy
A column of rectangular section b × d is subjected to an eccentric point load P with eccentricity e in the direction of the dimension d. The bending stress developed is:
A beam has an angle (L) section with one leg vertical and the other leg horizontal at the top. Which of the following cross-sections has the same nature of shear stress distribution as the angle section?
Shear stress depends on the width at each level, not on where that width sits. The angle and the T have the same width profile — a wide top leg over a narrow stem — so their diagrams match.
Question 5Easy
The stress distribution diagram at the base of a circular chimney is a rectangle (uniform compressive stress over the whole base). From this it may be concluded that:
Answer: C. the chimney is subjected to direct compressive stress only
A uniform (rectangular) diagram means the same stress everywhere, i.e. only direct compression from the self-weight, with no bending.
Question 6Medium
The shear stress distribution of a beam section is zero at the top, jumps suddenly to a higher value at one junction near the top, and then follows a parabolic curve down to zero at the bottom. Identify the beam section.
A single jump near the top means one wide top leg over a narrow stem — the angle section. Triangle and circle give smooth curves; the I has two jumps.
Question 7Easy
The maximum shear force in a rectangular beam section of depth 150 mm is 50 kN. If the permissible maximum shear stress is 5 MPa, the width of the section is:
τmax = 1.5V/(BD) ⇒ 5 = 1.5 × 50 000/(B × 150) ⇒ B = 100 mm.
Question 8Medium
A circular beam section is subjected to a shear force of 40π kN. The maximum shear stress the material can take is 6 MPa. With a factor of safety of 2, the safe diameter of the section is:
τ = (6V/bd³)(d²/4 − y²), a second-degree function of y: a parabola, zero at the top and bottom and maximum at the N.A.
Question 11Medium
A homogeneous simply supported prismatic beam of width B, depth D and span L carries a concentrated load of 200 kN, which can be placed anywhere on the span. The maximum flexural stress developed in the beam is:
Answer: D. a single symmetric parabola, maximum at mid-depth
τ = (6V/bd³)(d²/4 − y²): a symmetric parabola, zero at top and bottom, maximum (1.5τavg) at the neutral axis.
Question 15Medium
For a simply supported rectangular beam with a central load W, the maximum bending stress is equal to the direct tensile stress W/(bd) produced by the same load. The ratio of length to depth of the beam is: