Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Shear force & bending moment

Shear and moment: V = ΣF on one side, M = ΣF·d on one side
Bending stress: σ = M·y/I
Rectangle: I = bh³/12
Section modulus: W = I/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid: ȳ = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Beam deflection

Point load at midspan: δ = PL³/(48EI)
Uniform load, simply supported: δ = 5wL⁴/(384EI)
Cantilever, end load: δ = PL³/(3EI)
Cantilever, uniform load: δ = wL⁴/(8EI)
Superposition: δ = δ₁ + δ₂

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).

P = 100 kNd = 25 mmL = 2.5 mE = 100 GPa

ε = µε

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 95 kNsegment 1: Ø25 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A stepped solid steel shaft is fixed at A. Segment AB has diameter 75 mm and length 1.5 m; segment BC has diameter 40 mm and length 2 m. A torque of 400 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 400 N·md₁ = 75 mm, L₁ = 1.5 md₂ = 40 mm, L₂ = 2 m

τ_max = MPa

4.Torsion of shaftsHard

A solid steel shaft of diameter 40 mm is fixed at A. A torque of 200 N·m acts at B, 1.2 m from A, and a torque of 500 N·m in the opposite direction acts at the free end C, a further 3 m along. Find the maximum shear stress τ_max in the shaft.

d = 40 mmT_B = 200 N·mT_C = 500 N·mAB = 1.2 mBC = 3 m

τ_max = MPa

5.Shear force & bending momentMedium

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

10 kN @ 2 m40 kN @ 8 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

6.Shear force & bending momentHard

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

20 kN @ 12 m50 kN @ 4 m4 kN/m, 0–8 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

7.Centroid & second moment of areaMedium

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 100 mmt_f = 30 mmt_w = 20 mmh_w = 220 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 50 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 160 mmt_f = 25 mmt_w = 18 mmh_w = 160 mmM = 50 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −30 MPa, σᵧ = −20 MPa and τₓᵧ = 10 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = −30 MPaσ_y = −20 MPaτ_xy = 10 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −20 MPa, σᵧ = 110 MPa and τₓᵧ = −20 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −20 MPaσ_y = 110 MPaτ_xy = −20 MPaθ = 45°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 15 kN/mL = 5 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth h_req of the section.

w = 3 kN/mL = 3.5 mE = 11 GPa (timber)b = 100 mm

h_req = mm

13.Column bucklingMedium

A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 1.5 mE = 70 GPa (aluminium)d = 20 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 3 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 2037 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·25²/4 = 490.9 mm²

σ = P/A = 100·1000/490.9 = 203.7 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 203.7/100000 = 0.002037 = 2037 µε

Answer

ε = 2037 µε

2.Axial stress & strain

Answer: σ_max = 840 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 193.5 MPa

σ₂ = 840 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 840 MPa

Answer

σ_max = 840 MPa

3.Torsion of shafts

Answer: τ_max = 31.83 MPa

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Stress in each segment

T = 400 N·m = 400000 N·mm

τ_AB = 16·400000/(π·75³) = 4.829 MPa

τ_BC = 16·400000/(π·40³) = 31.83 MPa

Answer

τ_max = 31.83 MPa

4.Torsion of shafts

Answer: τ_max = 39.79 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 500 N·m

T_AB = T_C − T_B = 500 − 200 = 300 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 500 N·m

τ_max = 16·500000/(π·40³) = 39.79 MPa

Answer

τ_max = 39.79 MPa

5.Shear force & bending moment

Answer: E (x = 8 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 21.67 kN, R₂ = 28.33 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = 43.33, M_C = 66.67, M_D = 90, M_E = 113.3 kN·m, M_F = 56.67, M_G = 0

Answer

Mark E: M_E = 113.3 kN·m

6.Shear force & bending moment

Answer: C (x = 4 m)

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 31 kN, R₂ = 71 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = 54, M_C = 92 kN·m, M_D = 14, M_E = −80, M_F = −40, M_G = 0

Answer

Mark C: M_C = 92 kN·m

7.Centroid & second moment of area

Answer: I_x = 45.84 × 10⁶ mm⁴

Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 100·30 = 3000 mm², y_f = 220 + 30/2 = 235 mm

web: A_w = 20·220 = 4400 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3000·235 + 4400·110)/7400 = 160.7 mm

Parallel-axis theorem

flange: 100·30³/12 + 3000·(235 − 160.7)² = 1.68 × 10⁷ mm⁴

web: 20·220³/12 + 4400·(110 − 160.7)² = 2.905 × 10⁷ mm⁴

I_x = 1.68 × 10⁷ + 2.905 × 10⁷ = 4.584 × 10⁷ mm⁴ = 45.84 × 10⁶ mm⁴

Answer

I_x = 45.84 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 123.8 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 160·25 = 4000 mm², y_f = 160 + 25/2 = 172.5 mm

web: A_w = 18·160 = 2880 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4000·172.5 + 2880·80)/6880 = 133.8 mm

Parallel-axis theorem

flange: 160·25³/12 + 4000·(172.5 − 133.8)² = 6.206 × 10⁶ mm⁴

web: 18·160³/12 + 2880·(80 − 133.8)² = 1.447 × 10⁷ mm⁴

I_x = 6.206 × 10⁶ + 1.447 × 10⁷ = 2.068 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 185 − 133.8 = 51.22 mm

σ = M·y/I = 5 × 10⁷·51.22/2.068 × 10⁷ = 123.8 MPa

Answer

σ_top = 123.8 MPa

9.Mohr’s circle & stress transformation

Answer: θ_s = 13.28 °

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = 10/11.18 = 0.8944, sin 2θ_s = −(−5)/11.18 = 0.4472

2θ_s = atan2(5, 10) = 26.57°

Angle on the element

θ_s = 13.28°

check: θ_s − θ_p = 13.28 − 58.28 = −45°

Answer

θ_s = 13.28°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = 65 MPa

Transformation equation

τ_θ = −−65·1 + −20·0 = 65 MPa

Answer

τ_θ = 65 MPa

11.Beam deflection

Answer: δ_max = 48.44 mm

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

w = 15 N/mm, L = 5000 mm

EI = 2.52 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·15·5000⁴/(384·2.52 × 10¹²) = 48.44 mm

Answer

δ_max = 48.44 mm

12.Beam deflection

Answer: h_req = 187.4 mm

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

w = 3 N/mm, L = 3500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 3500/360 = 9.722 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·3·3500⁴/(384·11000·9.722) = 5.481 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·5.481 × 10⁷/100)^1/3 = 187.4 mm

Answer

h_req = 187.4 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 2.412 kN

Euler buckling: a slender column becomes unstable at P_cr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·7854/1500² = 2412 N = 2.412 kN

Answer

P_cr = 2.412 kN

14.Column buckling

Answer: σ_cr = 87.73 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/10 = 150

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/150² = 87.73 MPa

Answer

σ_cr = 87.73 MPa

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