Practice Questions

the strength of the beam will depend on the nature of loading

The most economical cross-section for a component subjected to bending is the I-section. The maximum stress depends on the section modulus for a given bending moment. The geometries that possess the highest section modulus are more economical among cross-sectional geometries with equivalent cross-sectional areas. The section modulus of the I-section is more than that of the rectangular cross-section for the same depth. The section modulus of the square cross- section is more than the circular cross-section.

For combined loading (bending and twisting); The equivalent torque is: $[T_{e}q = \sqrt{(M ^ 2 + T ^ 2)}]$ . Given: M = 3kNm and T = 4 kNm; Maximum torque applied is, $T_{e}q = \sqrt{(3 ^ 2 + 4 ^ 2)} = \sqrt{25} = 5kNm$

For a rectangular cross section beam, maximum shear stress, $\tau_{max} = \frac 32 \tau_ {Avg} \Rightarrow \frac {\tau_{max}} {\tau_{Avg}} = \frac 32$ = stress to average shear stress ratio of Maximum shear

When a load is applied at a distance from the axis of a column, it is called an eccentric load. Eccentric loading causes both direct stress and bending stress. In the case of a column under compressive load, direct compressive stress is uniform across the cross-section of the column. Eccentric loading applies to bending stress, not tensile or thermal stress.

The shear stress distribution for a rectangular cross-section beam subjected to maximum shear force F, $\tau= \frac {FA \hat y}{ bI} = \frac {F}{2I} (\frac {d ^ 2}{4}-  y ^ 2)$ . Shear stress is zero at the top and bottom fibre (y = d / 2) and maximum at neutral axis ( y = 0).

The bending stress in a beam varies linearly from zero at the neutral axis to a maximum at the outer surface. The neutral axis is subjected to zero bending stress as per the bending equation. The stress is directly proportional to the distance from the neutral axis.

For a round shaft under pure torsion, the shear stress is maximum at the outer surface and directly proportional to the distance from the center. At the center, the radius is zero, so the shear stress is zero. The torsion equation for a circular member is $\frac {\tau}{r} =\frac{T}{j} = \frac {C\theta}{I}$

Shear stress in a uniform I-section beam is given by f = VQ / lb The intensity of shear stress is maximum at the centroid of the section, which is the neutral axis. For a symmetric I-section, the shear stress is zero at the extreme fibres. Therefore, the largest shear stress acts at the neutral axis.

Shear stress is uniform throughout the cross section

The moment carrying capacity of a square cross-section beam of dimension D to the moment carrying capacity of a circular cross-section of diameter D is $\frac{16}{3\pi}$ This is obtained by calculating the section modulus of both shapes and taking their ratio. The MOI of circle and square cross-sections are given. Using the formula for section modulus and substituting the values, we get the ratio of moment carrying capacity.

The steel bars in a concrete beam are embedded near the bottom section. This is because the bottom part of the beam is subjected to tension and the steel bars are stronger in tension than compression. Concrete is stronger in compression and is embedded near the top section. Together, steel and concrete resist the downward load.

During transverse vibrations, the shaft experiences bending stresses as the particles move perpendicular to the axis. Longitudinal vibrations induce tensile and compressive stresses, while torsional vibrations induce torsional shear stresses. These stress types are dependent on the direction of particle movement.

The bending stress on the neutral axis of a cross-sectional beam is zero. This is because the layer of fibers in between is neither in tension nor compression. The neutral axis is an axis where there are no longitudinal stresses or strains. Shear stress is non-zero and mostly maximum along the neutral axis.

The intensity of bending stress in a beam varies directly with the distance of the point from the neutral axis. This is shown by the bending equation sigma = M/I * y where stress is proportional to the distance from the neutral axis. The stress varies linearly from zero at the neutral axis to a maximum at the outer surface of the beam.

The stress variation in the simple bending of beams is linear, as the bending stress varies linearly from zero at the neutral axis to a maximum at the outer surface.

Curved beams have a shifted neutral axis and non-linear, hyperbolic stress distribution when subjected to pure-bending couple. The bending stress is given by$ \sigma_{theta} = -\frac{My}{Ae(R - y)}$ where M is bending moment, y is distance axis, A is cross-sectional area, e is $R - R_{N}$, R is radius of centroidal axis and $R_{N}$ is radius of neutral axis.

The bending moment of a beam has its maximum value where the shear force is zero. This is because the rate of change of bending moment is equal to the shear force. The maximum and minimum values of bending moment occur where the slope of the bending moment distribution is zero.

The Sectional Modulus (Z) is the ratio of Moment of Inertia (I) of the beam cross-section about the neutral axis to the distance (ymax) of extreme fiber from the neutral axis.

The maximum compressive stress on a rectangular beam loaded vertically downwards occurs on the top layer. This is because where the stress is highest at the extreme fibres of the member.

In a curved beam with a symmetrical section, the maximum bending stress does not occur at the neutral axis or the centroidal axis. Instead, it is found at one of the fibres. The stress distribution is such that the stress is highest at the inside fibre, which is subjected to the maximum compressive stress, making it the critical point for stress analysis in such beams.

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