Questions: Mechanical Properties of Solids

Q1. Why does a solid push back when it is deformed, and what exactly is stress?

When a solid is deformed, its atoms move away from the positions where the interatomic forces balance. The material responds with an internal force that tries to bring the atoms back. The internal restoring force per unit area across a section is called stress. It is measured in N m^-2 or pascal. For a stretched wire carrying a load F, the tensile stress at a section well away from the ends is F/A. Stress describes the intensity of the internal force, not the force alone: the same load on a thinner wire produces a larger stress.

Hinglish: Jab solid deform hota hai to atoms apni balance position se hat jaate hain, aur interatomic force unhe wapas kheenchti hai. Us internal restoring force per unit area ko stress kehte hain. Wire ke liye stress F/A hai, isliye patla wire zyada stress jhelta hai.

Q2. Distinguish between longitudinal, shearing and hydraulic stress, and give the strain that goes with each.

Longitudinal stress acts along the length of a rod, either stretching or compressing it, and the matching strain is the change in length divided by the original length. Shearing stress acts parallel to a face while the opposite face is fixed, so layers slide; the matching strain is the sideways displacement divided by the perpendicular distance between the faces. Hydraulic stress acts normal to every face of a body immersed in a fluid, so the body shrinks uniformly; the matching strain is the change in volume divided by the original volume. Only the first kind changes the length along one direction strongly, the second changes shape without changing volume, and the third changes volume without changing shape.

Hinglish: Longitudinal stress length ke along hota hai aur length badalta hai. Shearing stress face ke parallel hota hai, layers slide karti hain, volume same rehta hai. Hydraulic stress chaar taraf se normal hota hai aur sirf volume ghatta hai, shape same rehta hai.

Q3. State Hooke's law, give its limits, and explain what the proportional limit and yield point mean.

Hooke's law states that while the deformation is small, the stress and the strain rise in proportion. The proportionality constant belongs to the material; for a wire pulled lengthwise it is Young's modulus. The relation survives only while the stress-strain graph is a straight line, that is up to the proportional limit. After that point the graph bends and the strain grows faster than the stress. The yield point is the stress at which permanent deformation begins, so a small extra stress produces a large extra strain. Unloading after yielding leaves a permanent set, because the atomic planes have slipped to new positions.

Hinglish: Hooke's law chhote deformation ke liye stress aur strain ko proportional kehta hai. Proportional limit ke baad curve mud jaata hai, aur yield point ke baad permanent deformation shuru hoti hai. Load hatane par bhi material purani length par nahi lauta.

Q4. A steel wire of length 2.0 m and diameter 1.0 mm hangs vertically and carries a 20 kg load. Find the stress, the strain and the extension. Take Y = 2.0 x 10^11 Pa.

The area is pi d^2/4 = pi (1.0 x 10^-3)^2/4 = 7.85 x 10^-7 m^2. The force is 20 x 9.8 = 196 N. Stress = 196/(7.85 x 10^-7) = 2.5 x 10^8 Pa. Strain = stress/Y = 2.5 x 10^8/(2.0 x 10^11) = 1.25 x 10^-3. Extension = strain x length = 1.25 x 10^-3 x 2.0 = 2.5 x 10^-3 m, that is 2.5 mm. This stress is below the yield strength of steel, so the wire behaves elastically.

Hinglish: Pehle area nikala, phir stress F/A, phir strain stress/Y, aur last me extension strain times length. Teen step me answer 2.5 mm aata hai, aur stress yield strength se kam hai, isliye wire elastic rehta hai.

Q5. Explain the terms ductile, brittle and elastomer using the stress-strain curve.

A ductile material, such as copper or mild steel, has a long plastic region between the yield point and the fracture point, so it can be drawn into wires without breaking. A brittle material, such as glass or cast iron, breaks soon after the proportional region, with very little plastic strain, and the curve ends near the elastic limit. An elastomer, such as rubber, has a long elastic region and can be stretched many times its original length, returning to its shape when the load is removed; its curve on unloading lies below the loading curve, forming a hysteresis loop. The area enclosed by the loop measures the energy lost as heat in each cycle.

Hinglish: Ductile material me plastic region lamba hota hai, isliye wire ban sakta hai. Brittle material proportional region ke baad jaldi toot jaata hai. Elastomer bahut zyada stretch hota hai aur load hatane par wapas aata hai, lekin loading aur unloading curves alag hote hain, isliye hysteresis loss hota hai.

Q6. What is the bulk modulus of water if a pressure of 2.0 x 10^6 Pa changes its volume by 0.09 percent?

The fractional change is 0.09/100 = 9.0 x 10^-4. The bulk modulus is B = P/(delta V/V) using magnitudes, so B = 2.0 x 10^6/(9.0 x 10^-4) = 2.2 x 10^9 Pa. The compressibility is the reciprocal, 1/B = 4.5 x 10^-10 Pa^-1. The value matches the standard figure for water, about 2.2 x 10^9 Pa, and it is many orders of magnitude larger than the bulk modulus of air at constant temperature.

Hinglish: Bulk modulus me pressure ko fractional volume change se divide karte hain. Yahan 0.09 percent ko 9.0 x 10^-4 likhna hai, phir P/(delta V/V) se 2.2 x 10^9 Pa milta hai, jo water ke liye standard value hai.

Q7. Why is the shear modulus of a liquid zero?

A liquid cannot sustain a shear stress at rest. When a tangential force is applied to a liquid, the layers slide past one another continuously and the liquid flows; there is no fixed sideways displacement that grows with the force and then stays. The shear modulus is defined as shear stress divided by shear strain, and because a liquid shows no elastic shear strain, the ratio is zero. The same reasoning applies to gases. Solids have a non-zero shear modulus because their atomic planes are held in place and oppose sliding.

Hinglish: Liquid par tangential force lagao to wo behne lagta hai, koi fixed elastic sideways strain nahi banti. Shear modulus stress aur strain ka ratio hai, aur strain zero hota hai, isliye G zero. Gases ka bhi G zero hota hai, solids ka nahi.

Q8. Derive the expression for the elastic potential energy stored in a stretched wire.

The force grows from zero to F as the extension grows from zero to delta L. If the material is elastic, the force is proportional to extension, so the work done is the area of the triangle under the force-extension graph, which is one half times F times delta L. Substituting F = Y A delta L/L gives U = (1/2) Y A (delta L)^2/L. Dividing by the volume A L gives the energy density u = (1/2) Y (delta L/L)^2 = (1/2) stress times strain. The same formula gives the energy stored per unit volume of any elastically deformed solid, provided the deformation is within the proportional region.

Hinglish: Force dheere dheere badhti hai, isliye work force-extension graph ke triangle ka area hai: half F delta L. Isko Y ke terms me likho aur volume se divide karo to energy density half stress times strain milti hai.

Q9. Two wires of the same material have lengths in the ratio 2:1 and diameters in the ratio 1:2. They carry the same load. Find the ratio of their extensions.

Extension is F L/(A Y). With F and Y the same, extension is proportional to L/A, and area is proportional to the square of the diameter. For wire 1, L/A = 2/(1)^2 in suitable units; for wire 2, L/A = 1/(2)^2 = 1/4. The ratio is 2 divided by 1/4 = 8, so the longer and thinner wire stretches eight times as much as the shorter and thicker wire. This is why long thin ropes and wires need much more care in structures.

Hinglish: Same material aur same load ke liye extension L/A ke proportional hota hai. Diameter square me aata hai, isliye lambi patli wire 8 guna zyada khinchti hai. Structure design me long thin members ka dhyaan rakhna padta hai.

Q10. On a stress-strain graph, material A has a steeper initial line and a lower peak than material B, whose curve is less steep but reaches a higher stress. Which is stiffer, and which is stronger?

The slope of the initial straight part gives Young's modulus, so the steeper material A is stiffer. The strength is the largest stress the material can bear before fracture, which is the peak of the curve, so material B is stronger. A roof beam needs stiffness to keep its shape under normal loads, while a crane hook needs strength to survive an overload. A material can be stiff without being strong, and the graph separates the two properties clearly.

Hinglish: Straight part ka slope stiffer material batata hai, isliye A stiffer hai. Curve ka peak strength batata hai, isliye B stronger hai. Stiffness aur strength do alag cheezein hain, aur graph par dono alag dikhte hain.

Quiz below: five questions on stress, the three moduli, elastic energy and the stress-strain curve. Try them without the formula sheet first, then check the explanations.