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

A bar is made of a steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 400 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 400 mm, E = 110 GPaδ = 0.25 mm

P = kN

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø120/Ø70 mm, E = 120 GPa, σ_allow = 60 MPa

P_max = kN

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 250 mm rotate 2.22° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 20 N·md = 15 mmL = 250 mmφ = 2.22°

G = GPa

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 50 mm is fixed at A. A torque of 1500 N·m acts at B, 2.5 m from A, and a torque of 100 N·m in the same direction acts at the free end C, a further 0.8 m along. Find the maximum shear stress τ_max in the shaft.

d = 50 mmT_B = 1500 N·mT_C = 100 N·mAB = 2.5 mBC = 0.8 m

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 55 kN·m at x = 4.5 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 6 mM₀ = 55 kN·m clockwise, at 4.5 m

M_x = kN·m

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment M_max in the beam.

L = 8 mw = 12 kN/m from 1 to 8 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 130 mm and inside diameter d = 106 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 130 mmd = 106 mm

r = mm

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 50 MPa in tension and 90 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 220 mmt_f = 10 mmt_w = 12 mmh_w = 280 mmσ_t,allow = 50 MPaσ_c,allow = 90 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −60 MPa, σᵧ = 60 MPa and τₓᵧ = 40 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −60 MPaσ_y = 60 MPaτ_xy = 40 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = −50 MPa and τₓᵧ = −30 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 = −70 MPaσ_y = −50 MPaτ_xy = −30 MPaσ_Y = 350 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 3 m from support A. Find the deflection δ_P under the load.

P = 19 kNa = 3 mL = 4 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the reaction R_B at the prop.

L = 8 mP = 70 kN at 4 m

R_B = kN

13.Column bucklingMedium

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

L = 5 mE = 70 GPa (aluminium)b × h = 40 × 100 mm

P_cr = kN

14.Column bucklingHard

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

L = 1 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 46.39 kN

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A₁ = 706.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·706.9) = 3.537 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(110000·1963) = 1.852 × 10⁻⁶ mm/N

Load

P = 0.25/5.389 × 10⁻⁶ = 46393 N = 46.39 kN

Answer

P = 46.39 kN

2.Axial stress & strain

Answer: P_max = 518.4 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·7461 = 1.037 × 10⁹ N

Governing strain

ε = min(0.0006, 0.0005) = 0.0005 (copper)

Largest load

P_max = 0.0005·1.037 × 10⁹ = 518363 N = 518.4 kN

Answer

P_max = 518.4 kN

3.Torsion of shafts

Answer: G = 25.96 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·15⁴/32 = 4970 mm⁴

φ = 2.22·π/180 = 0.03875 rad

Shear modulus

T = 20 N·m = 20000 N·mm

G = TL/(Jφ) = 20000·250/(4970·0.03875) = 25964 MPa

G = 25.96 GPa

Answer

G = 25.96 GPa

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

4.Torsion of shafts

Answer: τ_max = 65.19 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 100 N·m

T_AB = T_C + T_B = 100 + 1500 = 1600 N·m

Maximum shear stress

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

|T|_max = 1600 N·m

τ_max = 16·1.6 × 10⁶/(π·50³) = 65.19 MPa

Answer

τ_max = 65.19 MPa

5.Shear force & bending moment

Answer: M_x = 13.75 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 55/6 = 9.167 kN (upward)

R_A = −R_B = −9.167 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −9.167·4.5 + 55 = 13.75 kN·m

Answer

M_x = 13.75 kN·m

6.Shear force & bending moment

Answer: M_max = 93.02 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.

UDL resultant: W = w·ℓ = 12·7 = 84 kN, acting at x̄ = 4.5 m

ΣM_A = 0: R_B·8 = 12·7·4.5

R_B = 378/8 = 47.25 kN

ΣF_y = 0: R_A = 84 − 47.25 = 36.75 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 1) = 0

x₀ = 1 + 36.75/12 = 4.063 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 1)²/2 = 36.75·4.063 − 12·3.063²/2 = 93.02 kN·m

Answer

M_max = 93.02 kN·m

7.Centroid & second moment of area

Answer: r = 41.93 mm

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. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 4448 mm²

I = π(D⁴ − d⁴)/64 = 7.823 × 10⁶ mm⁴

Radius of gyration

r = √(I/A) = √(7.823 × 10⁶/4448) = 41.93 mm

Answer

r = 41.93 mm

8.Centroid & second moment of area

Answer: M_allow = 12.65 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 = (2200·285 + 3360·140)/5560 = 197.4 mm

top edge: c = 290 − 197.4 = 92.63 mm

Second moment of area

I = 220·10³/12 + 2200·87.63² + 12·280³/12 + 3360·57.37² = 4.992 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 90·4.992 × 10⁷/92.63 = 48.51 kN·m

bottom edge: M = 50·4.992 × 10⁷/197.4 = 12.65 kN·m

Governing value

M_allow = min = 12.65 kN·m (bottom edge)

Answer

M_allow = 12.65 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_max = 72.11 MPa

Radius of Mohr’s circle

τ_max = R = √(((−60 − 60)/2)² + 40²) = 72.11 MPa

Answer

τ_max = 72.11 MPa

10.Mohr’s circle & stress transformation

Answer: n = 3.82

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 = −60 MPa, R = 31.62 MPa

σ₁ = −28.38 MPa, σ₂ = −91.62 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 91.62 MPa

n = 350/91.62 = 3.82

Answer

n = 3.82

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

11.Beam deflection

Answer: δ_P = 1.272 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 = 19000 N, a = 3000 mm, b = 1000 mm, L = 4000 mm

EI = 1.12 × 10¹³ N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 19000·3000²·1000²/(3·1.12 × 10¹³·4000) = 1.272 mm

Answer

δ_P = 1.272 mm

12.Beam deflection

Answer: R_B = 21.88 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = 5PL³/(48EI) → R_B = 5P/16 = 5·70/16 = 21.88 kN

Answer

R_B = 21.88 kN

13.Column buckling

Answer: P_cr = 58.95 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·533333/2500² = 58954 N = 58.95 kN

Answer

P_cr = 58.95 kN

14.Column buckling

Answer: σ_cr = 69.09 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 = 1·1000 = 1000 mm

KL/r = 1000/10 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa

Answer

σ_cr = 69.09 MPa

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