Compound Stresses Questions and Answers

Answer: A

Both [a] and [r] are true, and [r] is the correct reason for [a].

Plane stress means the stresses on the faces normal to the thickness are zero: \(\sigma_{z} = \tau_{xz} = \tau_{yz} = 0\). With no out-of-plane shear stress, the shear strains \(\gamma_{xz}\) and \(\gamma_{yz}\) are zero. But the normal strain in the thickness direction is

\[ \varepsilon_{z} = -\frac{\nu}{E}(\sigma_{x} + \sigma_{y}) \]

which is not zero, so [a] is true. That strain arises because the in-plane stresses make the body contract or expand sideways, the Poisson effect, which is what [r] says, so [r] explains [a]. 'Both false' and '[r] false' are therefore wrong, and so is 'not the correct reason'.

Answer: D

The true statements are P and R.

Plane strain means the body deforms only in one plane, say x-y: the out-of-plane strains are zero, \(\varepsilon_{zz} = \gamma_{xz} = \gamma_{yz} = 0\). That is statement P.

Zero shear strains \(\gamma_{xz}\) and \(\gamma_{yz}\) give zero shear stresses \(\tau_{xz}\) and \(\tau_{yz}\). But stopping the normal strain \(\varepsilon_{zz}\) requires a normal stress: from Hooke's law, \(\varepsilon_{zz} = 0\) gives \(\sigma_{zz} = \nu(\sigma_{xx} + \sigma_{yy})\), which is generally not zero. That is statement R.

Statement Q, with \(\sigma_{zz}\) also zero, describes plane stress, the state of a thin plate, not plane strain. So 'Q only' and 'P and Q' are wrong, and 'P only' leaves out R, which is also true.

Answer: B

There is a constant body force per unit volume in the y-direction.

The equilibrium equations are \(\partial\sigma_{ij}/\partial x_{j} + b_{i} = 0\). With all shear stresses zero and the normal stresses depending only on y:

\[ \begin{aligned} x:\ \frac{\partial \sigma_{xx}}{\partial x} + b_{x} = 0 &\Rightarrow b_{x} = 0 \\ y:\ \frac{\partial \sigma_{yy}}{\partial y} + b_{y} = 0 &\Rightarrow b_{y} = -C_{2} \\ z:\ \frac{\partial \sigma_{zz}}{\partial z} + b_{z} = 0 &\Rightarrow b_{z} = 0 \end{aligned} \]

The body force is \(-C_{2}\) in the y-direction, constant and non-zero because \(C_{2} \neq 0\). It does not vary with y, and its direction is set by the sign of \(C_{2}\), not \(C_{1}\), since \(\sigma_{xx}\) and \(\sigma_{zz}\) vary only with y and contribute nothing to the x and z equations.

Answer: D

The stresses are 13.13 N/mm² (Compressive) and 1.88 N/mm² (Compressive).

The section is 100 mm along \(PQ\) and 80 mm across. The load is 30 mm from \(PQ\), so it is offset \(e = 40 - 30 = 10\) mm from the centroid towards \(PQ\), and centred along \(PQ\). Direct and bending stresses:

\[ \begin{aligned} \sigma_{d} &= \frac{60000}{100 \times 80} = 7.5\ \text{N/mm}^{2} \\ \sigma_{b} &= \frac{6Pe}{bh^{2}} = \frac{6 \times 60000 \times 10}{100 \times 80^{2}} \\ &= 5.625\ \text{N/mm}^{2} \end{aligned} \]

On face \(PQQ'P'\), nearer the load, the two add: \(7.5 + 5.625 = 13.13\) compressive. On face \(SRR'S'\) they subtract: \(7.5 - 5.625 = 1.88\), still compressive, because the load lies inside the middle third (the kern, \(e \le 80/6 \approx 13.3\) mm) and the section is nowhere in tension.

18.75 and 3.75 correspond to an eccentricity of 20 mm, twice the true offset, and the tensile options would need the load to lie outside the kern.