201. A vertical pillar supports a roof and is mainly under compression. If the load on the pillar is \(W\) and its cross-sectional area is \(A\), the average compressive stress is
ⓐ. \(\frac{W}{A}\)
ⓑ. \(\frac{A}{W}\)
ⓒ. \(WA\)
ⓓ. \(\frac{W}{A^2}\)
Correct Answer: \(\frac{W}{A}\)
Explanation: Compressive stress is normal stress produced when a body is squeezed or shortened. In a pillar supporting a roof, the roof load acts along the length of the pillar and tends to compress it. Average stress is force divided by cross-sectional area, so \(\sigma=\frac{W}{A}\). The same mathematical form is used for tensile and compressive normal stress, although the deformation direction is different. If a sign convention is used, compression may be assigned a negative sign, but the magnitude is still \(\frac{W}{A}\).
202. A square concrete pillar of side \(0.20\,\text{m}\) supports a load of \(8.0\times10^4\,\text{N}\). The average compressive stress in the pillar is
ⓐ. \(4.0\times10^5\,\text{Pa}\)
ⓑ. \(1.6\times10^4\,\text{Pa}\)
ⓒ. \(2.0\times10^7\,\text{Pa}\)
ⓓ. \(2.0\times10^6\,\text{Pa}\)
Correct Answer: \(2.0\times10^6\,\text{Pa}\)
Explanation: \( \textbf{Known data:} \) Load \(W=8.0\times10^4\,\text{N}\).
Side of square pillar \(s=0.20\,\text{m}\).
\( \textbf{Required quantity:} \) Compressive stress \(\sigma\).
Area of square cross-section is:
\[A=s^2\]
\[A=(0.20)^2=0.040\,\text{m}^2\]
Stress is:
\[\sigma=\frac{W}{A}\]
Substitute:
\[\sigma=\frac{8.0\times10^4}{0.040}\,\text{Pa}\]
\[\sigma=2.0\times10^6\,\text{Pa}\]
The side length must be squared before dividing the load by area.
\( \textbf{Final answer:} \) \(2.0\times10^6\,\text{Pa}\).
203. A long thin pillar and a short thick pillar are made of the same material and have the same cross-sectional area at the base. Under large compressive loads, the long thin pillar is more likely to fail by sideways bending. This qualitative failure is called
ⓐ. hydraulic compression
ⓑ. elastic hysteresis
ⓒ. lateral strain
ⓓ. buckling
Correct Answer: buckling
Explanation: A pillar under compression may not fail only because the average compressive stress exceeds the material strength. A long slender pillar can bend sideways under compression, a failure mode called buckling. This is why structural design depends not only on material and area but also on shape and length. A short thick pillar is generally more stable against sideways bending. Buckling is a structural stability issue, not simply a new unit or a different kind of strain.
204. The same compressive load is supported by two pillars of the same material. Pillar \(P\) has larger cross-sectional area than pillar \(Q\), while their lengths are similar. The safer statement is that pillar \(P\) has
ⓐ. smaller average compressive stress
ⓑ. larger average compressive stress
ⓒ. no compressive stress
ⓓ. the same stress regardless of area
Correct Answer: smaller average compressive stress
Explanation: Average compressive stress is \(\sigma=\frac{F}{A}\). If the same load \(F\) is carried by a larger area, the stress is smaller. This is one reason pillars and foundations are designed with sufficient cross-sectional area. It does not mean the pillar has no stress; it means the load is distributed over more area. Length and shape can also matter for stability, but the stress comparison here is controlled by area.
205. Match the structural situation with the most relevant mechanical idea.
| Situation | Mechanical idea |
|---|
| P. A roof load presses vertically on a pillar | 1. Compressive stress |
| Q. A slender column bends sideways under compression | 2. Buckling tendency |
| R. A cable lengthens under a hanging load | 3. Tensile strain |
The correct matching is
ⓐ. \(P-2\), \(Q-3\), \(R-1\)
ⓑ. \(P-3\), \(Q-1\), \(R-2\)
ⓒ. \(P-1\), \(Q-2\), \(R-3\)
ⓓ. \(P-1\), \(Q-3\), \(R-2\)
Correct Answer: \(P-1\), \(Q-2\), \(R-3\)
Explanation: A pillar carrying a roof is under compression, so the relevant stress is compressive stress. A slender column bending sideways under an axial compressive load shows buckling tendency. A cable lengthening under a hanging load undergoes tensile strain. These are all mechanical responses of solids, but they are not the same response. The loading direction and the observed deformation decide the correct classification.
206. A simply supported horizontal beam carries a load at its middle and bends downward. In the usual description of this bending, the upper layers of the beam are mainly
ⓐ. stretched, while the lower layers are compressed
ⓑ. free from all stress
ⓒ. changed only in volume without shape change
ⓓ. compressed, while the lower layers are stretched
Correct Answer: compressed, while the lower layers are stretched
Explanation: When a beam bends downward under a central load, its upper part becomes shorter along the curved length. This means the upper layers are mainly under compression. The lower part becomes longer along the curved length, so the lower layers are mainly under tension. Between these regions there is a layer whose length changes very little. Bending is therefore a combination of stretching and compression, not a separate deformation with no stress.
207. In a bent beam, the neutral layer is the layer that
ⓐ. has maximum tensile strain
ⓑ. has maximum compressive strain
ⓒ. does not stretch or compress
ⓓ. has only hydraulic pressure
Correct Answer: does not stretch or compress
Explanation: During bending, layers on one side of the beam are stretched and layers on the other side are compressed. The neutral layer lies between these stretched and compressed regions. Its length remains nearly unchanged during bending. This does not mean the whole beam is stress-free; it means this particular layer has zero longitudinal strain in the simplified picture. The idea helps explain why material placed far from the neutral layer can be very effective in resisting bending.
208. An \(I\)-shaped steel girder is often used in bridges and buildings because much of its material is placed far from the neutral layer. The main benefit is that it
ⓐ. removes all normal stress from the beam
ⓑ. makes Young’s modulus become zero
ⓒ. resists bending while saving material
ⓓ. turns bending into hydraulic compression
Correct Answer: resists bending while saving material
Explanation: In bending, the outer layers of a beam experience larger tensile and compressive strains than layers near the neutral layer. Placing more material far from the neutral layer helps the beam resist bending efficiently. An \(I\)-section achieves this by keeping material in the top and bottom flanges while reducing material near the middle. It does not eliminate stress completely; it distributes material where it is most useful. The explanation is qualitative here, so no advanced bending formula is needed.
209. Consider the following statements about bending of beams.
I. Bending involves both tensile and compressive effects.
II. A neutral layer has almost no change in length during bending.
III. In a downward-bent beam, upper and lower layers always have identical types of strain.
IV. Beam shape affects resistance to bending.
The suitable set is
ⓐ. I, II, and IV only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II, III, and IV
Correct Answer: I, II, and IV only
Explanation: Statement I is correct because bending combines stretching on one side and compression on the other. Statement II is correct because the neutral layer is the layer with nearly unchanged length. Statement III is not correct because the upper and lower layers of a bent beam usually have opposite types of strain. Statement IV is correct because structural shapes such as \(I\)-sections can improve resistance to bending. Bending cannot be understood by material alone; geometry also matters.
210. A metal rod is fixed rigidly between two supports. Its temperature is increased, but the supports prevent it from expanding. The rod develops stress mainly because
ⓐ. the rod’s mass becomes zero
ⓑ. strain becomes a pressure unit
ⓒ. the rod expands freely without opposition
ⓓ. supports prevent thermal expansion
Correct Answer: supports prevent thermal expansion
Explanation: A free rod expands when heated and contracts when cooled. If expansion is prevented by rigid supports, the rod cannot take its natural expanded length. The supports exert forces on the rod, and internal stress develops as a result. This is called thermal stress in a constrained body. The stress is not caused by a change in mass or by a change in the unit of strain; it comes from preventing the thermal deformation.
211. A wire is held between two rigid supports and cooled. If it is not allowed to contract, the wire tends to develop
ⓐ. hydraulic stress only
ⓑ. zero stress in all cases
ⓒ. shearing strain only
ⓓ. tensile stress
Correct Answer: tensile stress
Explanation: When a wire is cooled freely, it tends to contract. If rigid supports prevent this contraction, the wire is effectively stretched relative to the length it wants to have at the lower temperature. This produces tensile stress in the wire. The stress is not hydraulic because the wire is not being compressed uniformly from all sides. The sign of thermal stress depends on whether the constrained body is prevented from expanding or from contracting.
212. Assertion \(P\): Gaps are left between sections of railway tracks or bridge structures.
Reason \(Q\): Metals can expand on heating, and preventing expansion completely may produce large thermal stresses.
ⓐ. Both \(P\) and \(Q\) are true, and \(Q\) explains \(P\)
ⓑ. Both \(P\) and \(Q\) are true, but \(Q\) does not explain \(P\)
ⓒ. \(P\) is true, but \(Q\) is false
ⓓ. \(P\) is false, but \(Q\) is true
Correct Answer: Both \(P\) and \(Q\) are true, and \(Q\) explains \(P\)
Explanation: Metal structures expand when temperature rises and contract when temperature falls. If a long metal track or bridge member is prevented from expanding, large internal stress can develop. Expansion gaps give the structure space to change length with temperature. This reduces the risk of buckling, distortion, or excessive stress. The reason directly explains why such gaps are useful in real structures.
213. A rod of Young’s modulus \(Y\) and coefficient of linear expansion \(\alpha\) is heated by \(\Delta T\), but its expansion is completely prevented. If the formula \(\sigma=Y\alpha\Delta T\) is supplied, the thermal stress depends on
ⓐ. only the original length of the rod
ⓑ. only the mass of the rod
ⓒ. \(Y\), \(\alpha\), and \(\Delta T\)
ⓓ. only the colour of the rod
Correct Answer: \(Y\), \(\alpha\), and \(\Delta T\)
Explanation: The supplied relation is \(\sigma=Y\alpha\Delta T\). It shows that thermal stress increases when the material has a larger Young’s modulus \(Y\), a larger expansion coefficient \(\alpha\), or a larger temperature change \(\Delta T\). The original length does not appear in this simplified stress relation for complete constraint. Length would affect the free expansion \(\Delta L=\alpha L\Delta T\), but the constrained stress depends on prevented strain. The formula links thermal expansion with elastic resistance.
214. A steel rod with \(Y=2.0\times10^{11}\,\text{Pa}\) and \(\alpha=1.2\times10^{-5}\,\text{K}^{-1}\) is heated by \(50\,\text{K}\) while its expansion is completely prevented. Using \(\sigma=Y\alpha\Delta T\), the thermal stress is
ⓐ. \(2.4\times10^6\,\text{Pa}\)
ⓑ. \(1.2\times10^8\,\text{Pa}\)
ⓒ. \(8.3\times10^{-9}\,\text{Pa}\)
ⓓ. \(1.0\times10^{13}\,\text{Pa}\)
Correct Answer: \(1.2\times10^8\,\text{Pa}\)
Explanation: \( \textbf{Known data:} \) Young’s modulus \(Y=2.0\times10^{11}\,\text{Pa}\).
Coefficient of linear expansion \(\alpha=1.2\times10^{-5}\,\text{K}^{-1}\).
Temperature rise \(\Delta T=50\,\text{K}\).
\( \textbf{Supplied relation:} \)
\[\sigma=Y\alpha\Delta T\]
This relation applies when the thermal expansion is completely prevented.
Substitute:
\[\sigma=(2.0\times10^{11})(1.2\times10^{-5})(50)\,\text{Pa}\]
First multiply \(1.2\times10^{-5}\) by \(50\):
\[(1.2\times10^{-5})(50)=60\times10^{-5}=6.0\times10^{-4}\]
Now multiply by \(Y\):
\[\sigma=(2.0\times10^{11})(6.0\times10^{-4})\,\text{Pa}\]
\[\sigma=12.0\times10^7\,\text{Pa}\]
\[\sigma=1.2\times10^8\,\text{Pa}\]
The large value shows why constrained thermal expansion is important in structures.
\( \textbf{Final answer:} \) \(1.2\times10^8\,\text{Pa}\).
215. A long metal bar is allowed to slide at one end while the other end is fixed. Compared with a similar bar fixed rigidly at both ends, heating it produces
ⓐ. larger thermal stress because sliding removes deformation
ⓑ. the same thermal stress in every support condition
ⓒ. only shear stress with no length-change tendency
ⓓ. smaller stress because expansion is partly allowed
Correct Answer: smaller stress because expansion is partly allowed
Explanation: Thermal stress becomes large when natural expansion or contraction is prevented. If one end of a bar can slide, the bar can expand more freely when heated. This reduces the constraint and therefore reduces the stress developed. A bar fixed at both ends has much less freedom to change length, so the supports must exert larger forces. The key condition is not heating alone, but heating while deformation is restricted.
216. The following table describes three practical situations.
| Situation | Most relevant elastic idea |
|---|
| P. A crane cable must not exceed its safe working stress | 1. Factor of safety |
| Q. A bridge beam must resist bending efficiently | 2. Neutral layer and distribution of material |
| R. A railway track must allow expansion on hot days | 3. Thermal stress due to constrained expansion |
The correct matching is
ⓐ. \(P-2\), \(Q-3\), \(R-1\)
ⓑ. \(P-1\), \(Q-2\), \(R-3\)
ⓒ. \(P-3\), \(Q-1\), \(R-2\)
ⓓ. \(P-1\), \(Q-3\), \(R-2\)
Correct Answer: \(P-1\), \(Q-2\), \(R-3\)
Explanation: A crane cable is designed using working stress, breaking stress, and factor of safety. A bridge beam resists bending better when its material is placed effectively relative to the neutral layer. A railway track needs expansion gaps because constrained thermal expansion can produce large stress. These examples all involve mechanical properties of solids, but each uses a different part of the chapter. Matching the physical situation to the correct elastic idea prevents using one formula for every structural problem.
217. A wire is tested within its elastic limit. Its stress-strain graph and load-extension graph are both straight lines. The slope of the stress-strain graph and the slope of the load-extension graph respectively represent
ⓐ. \(\frac{AY}{L}\) and \(Y\)
ⓑ. work done and elastic energy density
ⓒ. \(Y\) and \(\frac{AY}{L}\)
ⓓ. strain and extension
Correct Answer: \(Y\) and \(\frac{AY}{L}\)
Explanation: On a stress-strain graph, the vertical quantity is stress and the horizontal quantity is strain. Therefore the slope is \(\frac{\text{stress}}{\text{strain}}\), which equals Young’s modulus \(Y\) in the linear elastic region. On a load-extension graph, the vertical quantity is load \(F\) and the horizontal quantity is extension \(\Delta L\). For a wire, \(\Delta L=\frac{FL}{AY}\), so \(\frac{F}{\Delta L}=\frac{AY}{L}\). The first slope is a material property, while the second also depends on the wire’s dimensions.
218. A load-extension graph for a wire is a straight line with slope \(2.0\times10^5\,\text{N m}^{-1}\). The wire has length \(1.0\,\text{m}\) and cross-sectional area \(5.0\times10^{-7}\,\text{m}^2\). Young’s modulus of the wire material is
ⓐ. \(1.0\times10^{-1}\,\text{Pa}\)
ⓑ. \(2.5\times10^{-12}\,\text{Pa}\)
ⓒ. \(1.0\times10^{5}\,\text{Pa}\)
ⓓ. \(4.0\times10^{11}\,\text{Pa}\)
Correct Answer: \(4.0\times10^{11}\,\text{Pa}\)
Explanation: \( \textbf{Known data:} \) Slope of load-extension graph:
\[k=\frac{F}{\Delta L}=2.0\times10^5\,\text{N m}^{-1}\]
Length:
\[L=1.0\,\text{m}\]
Area:
\[A=5.0\times10^{-7}\,\text{m}^2\]
For a wire in the elastic range:
\[k=\frac{AY}{L}\]
Rearranging:
\[Y=\frac{kL}{A}\]
Substitute:
\[Y=\frac{(2.0\times10^5)(1.0)}{5.0\times10^{-7}}\,\text{Pa}\]
\[Y=0.40\times10^{12}\,\text{Pa}\]
\[Y=4.0\times10^{11}\,\text{Pa}\]
The load-extension slope is not itself \(Y\); it becomes \(Y\) only after multiplying by \(\frac{L}{A}\).
\( \textbf{Final answer:} \) \(4.0\times10^{11}\,\text{Pa}\).
219. A force-extension graph for an elastic wire is a straight line from the origin to the point \(F=120\,\text{N}\), \(\Delta L=2.0\,\text{mm}\). The work done in stretching the wire up to this point is
ⓐ. \(0.24\,\text{J}\)
ⓑ. \(0.12\,\text{J}\)
ⓒ. \(120\,\text{J}\)
ⓓ. \(2.4\times10^{-3}\,\text{J}\)
Correct Answer: \(0.12\,\text{J}\)
Explanation: \( \textbf{Known data:} \) Final force \(F=120\,\text{N}\).
Final extension:
\[\Delta L=2.0\,\text{mm}=2.0\times10^{-3}\,\text{m}\]
\( \textbf{Required quantity:} \) Work done in stretching.
Work done is the area under the force-extension graph.
Since the graph is a straight line from the origin, the area is triangular:
\[W=\frac{1}{2}F\Delta L\]
Substitute:
\[W=\frac{1}{2}(120)(2.0\times10^{-3})\,\text{J}\]
\[W=60\times2.0\times10^{-3}\,\text{J}\]
\[W=120\times10^{-3}\,\text{J}\]
\[W=0.12\,\text{J}\]
The factor \(\frac{1}{2}\) is needed because the force rises gradually from \(0\) to \(120\,\text{N}\).
\( \textbf{Final answer:} \) \(0.12\,\text{J}\).
220. A stress-strain graph is linear from the origin to a point where stress is \(6.0\times10^7\,\text{Pa}\) and strain is \(3.0\times10^{-4}\). The area under the graph up to that point represents
ⓐ. \(1.8\times10^4\,\text{J m}^{-3}\)
ⓑ. \(9.0\times10^3\,\text{J m}^{-3}\)
ⓒ. \(2.0\times10^{11}\,\text{J m}^{-3}\)
ⓓ. \(5.0\times10^{-12}\,\text{J m}^{-3}\)
Correct Answer: \(9.0\times10^3\,\text{J m}^{-3}\)
Explanation: \( \textbf{Graph meaning:} \) Area under a stress-strain graph gives elastic energy density.
The graph is a straight line from the origin, so the area is triangular.
Stress at the point:
\[\sigma=6.0\times10^7\,\text{Pa}\]
Strain at the point:
\[\epsilon=3.0\times10^{-4}\]
Elastic energy density:
\[u=\frac{1}{2}\sigma\epsilon\]
Substitute:
\[u=\frac{1}{2}(6.0\times10^7)(3.0\times10^{-4})\]
\[(6.0\times10^7)(3.0\times10^{-4})=18.0\times10^3\]
\[u=9.0\times10^3\,\text{J m}^{-3}\]
The area gives energy per unit volume, while the slope of the same graph gives modulus.
\( \textbf{Final answer:} \) \(9.0\times10^3\,\text{J m}^{-3}\).
