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 elongation δ of the copper bar under the axial load.

P = 120 kNd = 20 mmL = 2 mE = 117 GPa

δ = mm

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 = 100 kNsegment 1: Ø30 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 0.5 m, copper (E = 117 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 10 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 400 mm rotate 11.97° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 20 N·md = 10 mmL = 400 mmφ = 11.97°

G = GPa

4.Torsion of shaftsHard

A solid steel shaft of diameter 20 mm is fixed at A. A torque of 200 N·m acts at B, 3 m from A, and a torque of 100 N·m in the opposite direction acts at the free end C, a further 2.5 m along. Find the maximum shear stress τ_max in the shaft.

d = 20 mmT_B = 200 N·mT_C = 100 N·mAB = 3 mBC = 2.5 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.

50 kN @ 3 m5 kN/m, 0–6 m

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

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 8.5 mw = 2 kN/m from 0 to 8.5 msupports at x = 0 and x = 6 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 190 mm wide and 18 mm thick, an overall depth of 320 mm and a web 12 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 190 mmH = 320 mmt_f = 18 mmt_w = 12 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 35 MPa in tension and 100 MPa in compression. Find the largest hogging bending moment M_allow the beam can carry (give its size).

b_f = 200 mmt_f = 20 mmt_w = 20 mmh_w = 140 mmσ_t,allow = 35 MPaσ_c,allow = 100 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 90 MPa, σᵧ = −60 MPa and τₓᵧ = −60 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 90 MPaσ_y = −60 MPaτ_xy = −60 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 20 MPa, σᵧ = −60 MPa and τₓᵧ = −40 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 20 MPaσ_y = −60 MPaτ_xy = −40 MPaσ_Y = 350 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δ_P under the load.

P = 20 kNa = 1 mL = 4 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 2 kNw = 2 kN/mL = 2.5 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 3 mE = 200 GPa (steel)b × h = 25 × 100 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 25 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 25 kNn = 2.5L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 6.529 mm

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·20²/4 = 314.2 mm²

σ = P/A = 120·1000/314.2 = 382 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 382/117000 = 0.003265 = 3265 µε

Elongation

δ = ε·L = 0.003265·2000 mm = 6.529 mm

Answer

δ = 6.529 mm

2.Axial stress & strain

Answer: σ_max = 497.4 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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 141.5 MPa

σ₂ = 497.4 MPa

Largest stress

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

σ_max = 497.4 MPa

Answer

σ_max = 497.4 MPa

3.Torsion of shafts

Answer: G = 39 GPa

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.

Polar moment and angle

J = π·10⁴/32 = 981.7 mm⁴

φ = 11.97·π/180 = 0.2089 rad

Shear modulus

T = 20 N·m = 20000 N·mm

G = TL/(Jφ) = 20000·400/(981.7·0.2089) = 39005 MPa

G = 39 GPa

Answer

G = 39 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: τ_max = 63.66 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 100 N·m

T_AB = T_C − T_B = 100 − 200 = −100 N·m

Maximum shear stress

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

|T|_max = 100 N·m

τ_max = 16·100000/(π·20³) = 63.66 MPa

Answer

τ_max = 63.66 MPa

5.Shear force & bending moment

Answer: D (x = 3 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 = 40 kN, R₂ = 40 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 = 37.5, M_C = 70, M_D = 97.5 kN·m, M_E = 70, M_F = 37.5, M_G = 0

Answer

Mark D: M_D = 97.5 kN·m

6.Shear force & bending moment

Answer: M_max = 6.146 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam, including the load on the overhang.

UDL resultant: W = w·ℓ = 2·8.5 = 17 kN, acting at x̄ = 4.25 m

ΣM_A = 0: R_B·6 = 2·8.5·4.25

R_B = 72.25/6 = 12.04 kN

ΣF_y = 0: R_A = 17 − 12.04 = 4.958 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 4.958/2 = 2.479 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 4.958²/(2·2) = 6.146 kN·m

Answer

M_max = 6.146 kN·m

7.Centroid & second moment of area

Answer: I_x = 179 × 10⁶ mm⁴

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 320 − 2·18 = 284 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (190·320³ − 178·284³)/12

I_x = 1.79 × 10⁸ mm⁴ = 179 × 10⁶ mm⁴

Answer

I_x = 179 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 12.43 kN·m

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.

Centroid

ȳ = (4000·150 + 2800·70)/6800 = 117.1 mm

top edge: c = 160 − 117.1 = 42.94 mm

Second moment of area

I = 200·20³/12 + 4000·32.94² + 20·140³/12 + 2800·47.06² = 1.525 × 10⁷ mm⁴

Moment allowed by each edge

Hogging: the top is in tension, the bottom in compression.

top edge: M = 35·1.525 × 10⁷/42.94 = 12.43 kN·m

bottom edge: M = 100·1.525 × 10⁷/117.1 = 13.03 kN·m

Governing value

M_allow = min = 12.43 kN·m (top edge)

Answer

M_allow = 12.43 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_max = 96.05 MPa

Radius of Mohr’s circle

τ_max = R = √(((90 − −60)/2)² + −60²) = 96.05 MPa

Answer

τ_max = 96.05 MPa

10.Mohr’s circle & stress transformation

Answer: n = 3.094

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = −20 MPa, R = 56.57 MPa

σ₁ = 36.57 MPa, σ₂ = −76.57 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 113.1 MPa

n = 350/113.1 = 3.094

Answer

n = 3.094

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ_P = 3.261 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.

P = 20000 N, a = 1000 mm, b = 3000 mm, L = 4000 mm

EI = 4.6 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 20000·1000²·3000²/(3·4.6 × 10¹²·4000) = 3.261 mm

Answer

δ_P = 3.261 mm

12.Beam deflection

Answer: δ_max = 2.661 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.

P = 2000 N, w = 2 N/mm, L = 2500 mm, EI = 6.27 × 10¹¹ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 2000·2500³/(48·6.27 × 10¹¹) = 1.038 mm

δ_w = 5wL⁴/(384EI) = 5·2·2500⁴/(384·6.27 × 10¹¹) = 1.622 mm

Add them

δ_max = 1.038 + 1.622 = 2.661 mm

Answer

δ_max = 2.661 mm

13.Column buckling

Answer: P_cr = 7.139 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 100·25³/12 = 130208 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·130208/6000² = 7139 N = 7.139 kN

Answer

P_cr = 7.139 kN

14.Column buckling

Answer: d = 49.09 mm

Required critical load

P_cr = n·P = 2.5·25000 = 62500 N

KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·62500·3000²/(π³·200000))^(1/4) = 49.09 mm

Answer

d = 49.09 mm

Check: the slenderness KL/r is 244.5, so the column is slender and Euler applies.

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