Diploma Civil and Mechanical Engineering for Engineering Assistant · Chapter 1

Strength of Materials

What to remember

Stress = load ÷ area, strain = change in length ÷ original length, and E = stress ÷ strain within the elastic limit (Hooke's law).
Beams bend by M/I = σ/y = E/R, shafts twist by T/J = τ/r = Gθ/L, and long columns buckle by Euler's formula Pcr = π²EI/Le².
Elastic constants are linked: E = 2G(1 + ν) = 3K(1 − 2ν). Factor of safety = ultimate stress ÷ working stress.

Simple stress and strain

Stress (σ) is internal resistance per unit area, σ = P/A. Unit: N/mm² = MPa (1 N/mm² = 1 MPa; 1 GPa = 1000 MPa).
Tensile and compressive stress act perpendicular to the area. Shear stress (τ) acts parallel to the area, τ = P/A.
Strain (ε) = change in length ÷ original length. It has no unit. Shear strain (γ) is the angular distortion in radians.
Hooke's law: stress is proportional to strain within the proportional (elastic) limit. E (Young's modulus) = σ/ε. For mild steel E is about 200 GPa.
Modulus of rigidity (G) = shear stress ÷ shear strain. Bulk modulus (K) = pressure ÷ volumetric strain.
Poisson's ratio (ν) = lateral strain ÷ longitudinal strain. For most metals it is about 0.25 to 0.33. Its theoretical upper limit is 0.5.
Extension of a bar: δ = PL/AE.
Volumetric strain of a bar under axial strain ε is ε(1 − 2ν).
Thermal stress in a fully restrained bar: σ = αΔT·E, where α is the coefficient of linear expansion. If free to expand there is no stress.
Factor of safety = ultimate stress ÷ working (allowable) stress. For ductile materials the yield (or 0.2% proof) stress is often used instead of the ultimate stress, as per the code; brittle materials use the ultimate stress.
Worked example 1: a 20 mm × 10 mm bar carries 40 kN. A = 200 mm², σ = 40,000/200 = 200 N/mm².
Worked example 2: P = 20 kN, L = 2 m, A = 200 mm², E = 200 GPa. δ = (20,000 × 2000) ÷ (200 × 200,000) = 1 mm.
Worked example 3: α = 12 × 10⁻⁶ per °C, ΔT = 50 °C, E = 200 GPa fully restrained. σ = 12 × 10⁻⁶ × 50 × 200,000 = 120 N/mm².

Stress-strain behaviour

For mild steel in tension, the curve has these points in order: proportional limit, elastic limit, upper and lower yield points, strain hardening, ultimate stress, breaking (fracture) stress. A neck forms after the ultimate stress.

Ductile materials (mild steel, copper) show large plastic deformation before failure. Brittle materials (cast iron, concrete, glass) fail with little strain.
Cold-worked or high-strength steels have no clear yield point, so a 0.2% proof stress is used.
Resilience is the strain energy stored in a body. Proof resilience is the energy stored at the elastic limit. Toughness is the total energy to fracture. Hardness is resistance to indentation or scratching.
Strain energy U = σ²/(2E) × volume. A load applied suddenly produces double the stress of the same load applied gradually.

Property	Meaning	Example
Elasticity	Returns to shape after load is removed	Steel spring
Plasticity	Permanent deformation	Clay
Ductility	Drawn into wire	Copper
Malleability	Hammered into sheets	Gold, lead
Brittleness	Breaks without warning	Cast iron

Elastic constants and compound bars

E = 2G(1 + ν) and E = 3K(1 − 2ν). Also E = 9KG ÷ (3K + G).
Worked example 4: E = 200 GPa and ν = 0.25. G = 200 ÷ (2 × 1.25) = 80 GPa. K = 200 ÷ (3 × 0.5) = 133.3 GPa.
Bars in series carry the same load, and total extension is the sum of the extensions of each part.
Bars in parallel (composite bars) have the same extension. The load is shared in proportion to AE. Equal strain in both materials means stress is in the ratio of their E values.

Beams: shear force and bending moment

Shear force (SF) is the algebraic sum of vertical forces on one side of a section. Bending moment (BM) is the algebraic sum of moments of forces on one side.
Relations: slope of the SF diagram = load intensity; slope of the BM diagram = shear force. Maximum BM occurs where SF is zero or changes sign.
A point of contraflexure is where BM changes sign.

Beam and load	Max BM	Max deflection
Cantilever, point load W at free end	WL (at fixed end)	WL³ ÷ 3EI
Cantilever, UDL w per unit length	wL² ÷ 2	wL⁴ ÷ 8EI
Simply supported, central point load W	WL ÷ 4	WL³ ÷ 48EI
Simply supported, UDL w	wL² ÷ 8	5wL⁴ ÷ 384EI

Shape of diagrams: for a UDL, SF is a sloping straight line and BM is a parabola. For a point load, SF is constant between loads and BM is a straight line.
For a simply supported beam with a central point load, each reaction is W/2.

Bending stress in beams

Bending equation: M/I = σ/y = E/R. Here M is the moment, I the moment of inertia of the section, y the distance from the neutral axis and R the radius of curvature.
Section modulus Z = I/ymax. Maximum bending stress = M/Z. A larger Z means a stronger section.
Neutral axis passes through the centroid. Stress is zero at the neutral axis and maximum at the extreme fibres.
Moment of inertia: rectangle I = bd³/12 (Z = bd²/6); solid circle I = πd⁴/64 (Z = πd³/32); hollow circle I = π(D⁴ − d⁴)/64.
Shear stress in beams: τ = V·A·ȳ ÷ (I·b). For a rectangular section maximum τ = 1.5 × average and occurs at the neutral axis. For a circular section maximum τ = 4/3 × average.
Worked example 5: a simply supported beam of 4 m span, UDL 10 kN/m, section 100 mm wide × 200 mm deep. M = 10 × 16 ÷ 8 = 20 kN·m. Z = 100 × 200² ÷ 6 = 666,667 mm³. σ = 20 × 10⁶ ÷ 666,667 = 30 N/mm².

Torsion of shafts

Torsion equation: T/J = τ/r = Gθ/L. J is the polar moment of inertia, θ the angle of twist in radians.
Solid shaft J = πd⁴/32. Hollow shaft J = π(D⁴ − d⁴)/32. Torsional shear stress is maximum at the outer surface and zero at the axis.
Power transmitted: P = 2πNT/60 watts, where N is in rpm and T in N·m.
Worked example 6: T = 1000 N·m at N = 300 rpm. P = 2π × 300 × 1000 ÷ 60 = 31,416 W, about 31.4 kW.
For the same weight, a hollow shaft transmits more torque than a solid shaft. Torsional rigidity is GJ.

Thin cylinders, columns and combined stress

Thin cylinder (t < d/20) with internal pressure p: hoop (circumferential) stress σc = pd ÷ 2t; longitudinal stress σl = pd ÷ 4t. Hoop stress is twice the longitudinal stress, so a boiler shell fails along its length first. A thin sphere has σ = pd ÷ 4t in every direction.
Worked example 7: p = 2 N/mm², d = 1000 mm, t = 10 mm. σc = 2 × 1000 ÷ 20 = 100 N/mm². σl = 50 N/mm².
Column: a long compression member. Euler's critical load Pcr = π²EI ÷ Le², where Le is the effective length.

End condition	Effective length Le
Both ends hinged	L
Both ends fixed	L/2
One end fixed, one free	2L
One end fixed, one hinged	L/√2

Slenderness ratio = Le/k, where k = √(I/A) is the radius of gyration. Euler's formula applies only to long columns with a high slenderness ratio. Short columns fail by crushing.
Principal stresses (on planes with no shear): σ1,2 = (σx + σy)/2 ± √[((σx − σy)/2)² + τ²]. Maximum shear stress = (σ1 − σ2)/2, which is the radius of Mohr's circle.

Exam traps

Stress is force per area (N/mm²). Strain has no unit.
Hoop stress (pd/2t) is double longitudinal stress (pd/4t). Do not swap them.
Max BM of a cantilever with end load is WL; for simply supported central load it is WL/4.
For a rectangle I = bd³/12, but for a circle I = πd⁴/64 and J = πd⁴/32. J is twice I for a circle.
Hinged-hinged Le = L, fixed-fixed Le = L/2 (not 2L). Fixed-free is 2L.
Max shear stress in a beam is 1.5 times the average for a rectangle but 4/3 times for a circle.
Free thermal expansion creates no stress. Stress arises only when expansion is prevented.
Suddenly applied load gives twice the stress of the same gradually applied load.

One-liners

1. 1 MPa = 1 N/mm².
2. Young's modulus of mild steel is about 200 GPa.
3. Poisson's ratio of metals is about 0.25 to 0.33.
4. E = 2G(1 + ν) = 3K(1 − 2ν).
5. Extension of a bar δ = PL/AE.
6. Thermal stress = αΔT·E for a restrained bar.
7. Maximum bending moment occurs where shear force is zero.
8. Bending equation M/I = σ/y = E/R.
9. Section modulus Z = I/y; rectangle Z = bd²/6.
10. Torsion equation T/J = τ/r = Gθ/L.
11. Power in a shaft P = 2πNT/60.
12. Euler's load Pcr = π²EI/Le², and a fixed-fixed column carries four times the load of a hinged-hinged one.

Practice questions

Stress is defined as
1. change in length per unit length
2. energy per unit volume
3. load per unit area
4. load per unit length
Answer
C. load per unit area
Stress = force ÷ area, measured in N/mm² (MPa).
The unit of strain is
1. Pascal
2. N/m
3. N/mm²
4. no unit (dimensionless)
Answer
D. no unit (dimensionless)
Strain is a ratio of two lengths.
Hooke's law holds good up to the
1. ultimate stress
2. proportional limit
3. breaking point
4. upper yield point
Answer
B. proportional limit
Stress is proportional to strain only within the proportional (elastic) limit.
Poisson's ratio is the ratio of
1. lateral strain to longitudinal strain
2. shear stress to shear strain
3. volumetric strain to linear strain
4. longitudinal strain to lateral strain
Answer
A. lateral strain to longitudinal strain
It compares sideways contraction with the lengthwise extension.
The modulus of rigidity is the ratio of
1. pressure to volumetric strain
2. shear stress to shear strain
3. lateral strain to linear strain
4. direct stress to direct strain
Answer
B. shear stress to shear strain
G = τ/γ.
Which material is brittle?
1. Cast iron
2. Aluminium
3. Mild steel
4. Copper
Answer
A. Cast iron
Cast iron fails with very little plastic strain.
Factor of safety is the ratio of
1. yield stress to strain
2. working stress to ultimate stress
3. proof load to area
4. ultimate stress to working stress
Answer
D. ultimate stress to working stress
FOS = ultimate stress ÷ working (allowable) stress.
The neutral axis of a beam passes through the
1. bottom fibre
2. top fibre
3. centroid of the section
4. support
Answer
C. centroid of the section
Bending stress is zero at the neutral axis, which passes through the centroid.
The moment of inertia of a rectangular section about its centroidal axis parallel to the base is
1. bd²/6
2. bd³/3
3. bd³/12
4. b³d/3
Answer
C. bd³/12
For a rectangle I = bd³/12.
The torsion equation is
1. T = Pd
2. T/J = τ/r = Gθ/L
3. M/I = σ/y = E/R
4. P/A = σ
Answer
B. T/J = τ/r = Gθ/L
This is the standard relation for twisting of a shaft.
The relation between slope of the bending moment diagram and shear force is that the
1. BM is always zero where SF is maximum
2. BM diagram equals the SF diagram
3. slope of the SF diagram equals the bending moment
4. slope of the BM diagram equals the shear force
Answer
D. slope of the BM diagram equals the shear force
dM/dx = V.
Euler's formula gives the buckling load of a
1. long column
2. short column
3. thin plate in tension
4. beam in bending
Answer
A. long column
Euler's formula is valid for long, slender columns.
The effective length of a column fixed at both ends is
1. L/√2
2. L
3. L/2
4. 2L
Answer
C. L/2
With both ends fixed the effective length is half the actual length.
The ratio of hoop stress to longitudinal stress in a thin cylindrical shell is
1. 0.5
2. 1
3. 4
4. 2
Answer
D. 2
σc = pd/2t and σl = pd/4t, so the ratio is 2.
The point at which bending moment changes sign is called
1. point of contraflexure
2. centre of gravity
3. neutral point
4. yield point
Answer
A. point of contraflexure
At the point of contraflexure the beam changes from sagging to hogging.
Maximum bending moment in a cantilever of length L with a point load W at its free end is
1. WL/4
2. WL
3. WL/8
4. WL/2
Answer
B. WL
The moment is greatest at the fixed support.
Maximum bending moment for a simply supported beam with a UDL w and span L is
1. wL²/8
2. wL²/4
3. wL²/2
4. wL/4
Answer
A. wL²/8
At mid span M = wL²/8.
A bar of cross-section 500 mm² carries a tensile load of 100 kN. The stress is
1. 20 N/mm²
2. 50 N/mm²
3. 500 N/mm²
4. 200 N/mm²
Answer
D. 200 N/mm²
σ = 100,000 ÷ 500 = 200 N/mm².
A rod of length 1000 mm extends by 2 mm. The strain is
1. 2.0
2. 0.0002
3. 0.002
4. 0.02
Answer
C. 0.002
ε = 2 ÷ 1000 = 0.002.
A rod has σ = 100 N/mm² and E = 200 GPa. The strain is
1. 0.0001
2. 0.0005
3. 0.005
4. 0.002
Answer
B. 0.0005
ε = 100 ÷ 200,000 = 0.0005.
A bar of 250 mm² area and 1 m length carries 50 kN. With E = 200 GPa, the extension is
1. 0.5 mm
2. 10 mm
3. 2 mm
4. 1 mm
Answer
D. 1 mm
δ = PL/AE = 50,000 × 1000 ÷ (250 × 200,000) = 1 mm.
For E = 200 GPa and ν = 0.25 the modulus of rigidity is
1. 100 GPa
2. 66.7 GPa
3. 80 GPa
4. 120 GPa
Answer
C. 80 GPa
G = E ÷ 2(1 + ν) = 200 ÷ 2.5 = 80 GPa.
For E = 150 GPa and ν = 1/3, the bulk modulus is
1. 150 GPa
2. 50 GPa
3. 450 GPa
4. 75 GPa
Answer
A. 150 GPa
K = E ÷ 3(1 − 2ν) = 150 ÷ (3 × 1/3) = 150 GPa.
A steel bar with α = 12 × 10⁻⁶/°C is fully restrained and heated by 40 °C (E = 200 GPa). The thermal stress is
1. 240 N/mm²
2. 96 N/mm²
3. 48 N/mm²
4. 120 N/mm²
Answer
B. 96 N/mm²
σ = 12 × 10⁻⁶ × 40 × 200,000 = 96 N/mm².
A simply supported beam of 6 m span carries a central point load of 12 kN. The maximum bending moment is
1. 9 kN·m
2. 18 kN·m
3. 36 kN·m
4. 72 kN·m
Answer
B. 18 kN·m
M = WL/4 = 12 × 6 ÷ 4 = 18 kN·m.
A 2 m cantilever carries a UDL of 5 kN/m over its whole length. The maximum bending moment is
1. 2.5 kN·m
2. 20 kN·m
3. 5 kN·m
4. 10 kN·m
Answer
D. 10 kN·m
M = wL²/2 = 5 × 4 ÷ 2 = 10 kN·m.
The section modulus of a rectangular beam 100 mm wide and 300 mm deep is
1. 1.5 × 10⁶ mm³
2. 3 × 10⁶ mm³
3. 0.5 × 10⁶ mm³
4. 4.5 × 10⁶ mm³
Answer
A. 1.5 × 10⁶ mm³
Z = bd²/6 = 100 × 90,000 ÷ 6 = 1.5 × 10⁶ mm³.
A beam with section modulus 500,000 mm³ carries a moment of 15 kN·m. The maximum bending stress is
1. 3 N/mm²
2. 7.5 N/mm²
3. 30 N/mm²
4. 75 N/mm²
Answer
C. 30 N/mm²
σ = 15 × 10⁶ ÷ 500,000 = 30 N/mm².
A thin cylinder with p = 1.5 N/mm², d = 800 mm and t = 8 mm has a longitudinal stress of
1. 150 N/mm²
2. 75 N/mm²
3. 37.5 N/mm²
4. 18.75 N/mm²
Answer
C. 37.5 N/mm²
σl = pd/4t = 1.5 × 800 ÷ 32 = 37.5 N/mm².
A shaft transmits a torque of 1000 N·m at 60 rpm. The power is about
1. 62.8 kW
2. 6.28 kW
3. 3.14 kW
4. 0.628 kW
Answer
B. 6.28 kW
P = 2πNT/60 = 2π × 60 × 1000 ÷ 60 = 6283 W.
A material has ultimate stress 400 N/mm² and a factor of safety of 4. The allowable stress is
1. 200 N/mm²
2. 400 N/mm²
3. 1600 N/mm²
4. 100 N/mm²
Answer
D. 100 N/mm²
Allowable stress = 400 ÷ 4 = 100 N/mm².
For equal E, I and length, the ratio of the Euler load of a fixed-fixed column to that of a hinged-hinged column is
1. 2
2. 1/4
3. 4
4. 1
Answer
C. 4
Pcr varies as 1/Le². Le is L/2 versus L, giving a ratio of 4.
The maximum shear stress in a rectangular beam section is related to the average shear stress as
1. 1.5 times
2. 2 times
3. 4/3 times
4. equal
Answer
A. 1.5 times
The parabolic shear distribution gives τmax = 1.5 τavg at the neutral axis.
A suddenly applied load, compared with the same load applied gradually, produces a stress that is
1. the same
2. twice as much
3. half as much
4. four times as much
Answer
B. twice as much
By energy balance, the sudden-load stress is double the gradual-load stress.
Which of the following has the largest deflection under a point load W at the free end for the same L, E and I?
1. Simply supported beam with UDL of total load W
2. Simply supported beam with central load (WL³/48EI)
3. Fixed beam with central load
4. Cantilever (WL³/3EI)
Answer
D. Cantilever (WL³/3EI)
The cantilever formula has the smallest denominator, so it deflects most.
Which statements are correct? 1. Hoop stress in a thin cylinder is pd/2t. 2. Longitudinal stress is pd/2t.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Longitudinal stress is pd/4t, not pd/2t.
Which statements are correct? 1. Bending stress is zero at the neutral axis. 2. Shear stress in a rectangular beam is zero at the neutral axis.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Shear stress is maximum at the neutral axis for a rectangular section.
Which statements are correct? 1. Maximum bending moment occurs where the shear force is zero. 2. Euler's formula is suitable for long columns.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are standard facts.
Which statements are correct? 1. Torsional shear stress is maximum at the axis of a shaft. 2. A hollow shaft is lighter than a solid shaft of equal strength.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Torsional stress is maximum at the outer surface, so statement 1 is wrong.
Which statements are correct? 1. A freely expanding bar develops thermal stress equal to αΔT·E. 2. A bar of rubber has a higher Young's modulus than steel.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
A free bar has no thermal stress, and steel is far stiffer than rubber.
Match: (a) UDL on a simply supported beam (b) Point load on a cantilever (c) Both ends hinged column (d) Thin sphere. Which pairing is correct?
1. (a) wL²/8, (d) pd/4t
2. (a) WL/4, (d) pd/2t
3. (b) WL/4, (c) Le = 2L
4. (a) wL²/2, (c) Le = L/2
Answer
A. (a) wL²/8, (d) pd/4t
The simply supported UDL moment is wL²/8, and a thin sphere has σ = pd/4t.
A solid circular shaft of diameter d has J = πd⁴/32. Doubling the diameter changes its torsional rigidity (GJ) by a factor of
1. 16
2. 8
3. 4
4. 2
Answer
A. 16
J varies as d⁴, so 2⁴ = 16.
A bar of a given length is loaded axially so that its cross-section area is doubled. For the same load, the extension becomes
1. one-fourth
2. double
3. unchanged
4. half
Answer
D. half
δ = PL/AE varies inversely with A.
A rectangular beam of width b and depth d is replaced by one of the same width but double depth. For the same moment, the bending stress becomes
1. half
2. one-fourth
3. one-eighth
4. double
Answer
B. one-fourth
Z = bd²/6 increases four times, so σ = M/Z falls to one-fourth.
The reaction at each support of a simply supported beam with a central point load of 20 kN is
1. 20 kN
2. 40 kN
3. 10 kN
4. 5 kN
Answer
C. 10 kN
By symmetry each support carries W/2.

Page 1 of 1

Finished reading?

Test yourself in 10 minutes

A free 10-question mini-mock on your exam's real pattern, with negative marking and an instant score. अपनी तैयारी 10 मिनट में जाँचें।

Take a free mini-mock →Topic-wise practice