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 = 95 kNd = 25 mmL = 2 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 total elongation δ.

P = 20 kNsegment 1: Ø30 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, steel (E = 200 GPa)

δ = mm

3.Torsion of shaftsMedium

A hollow steel shaft (outer diameter 30 mm, inner diameter 21 mm, length 1.5 m) is fixed at one end and twisted by a torque of 250 N·m at the other. Find the angle of twist φ in degrees.

T = 250 N·md_o = 30 mmd_i = 21 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 30 mm is fixed at A. A torque of 100 N·m acts at B, 1.2 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 2 m along. Find the maximum shear stress τ_max in the shaft.

d = 30 mmT_B = 100 N·mT_C = 200 N·mAB = 1.2 mBC = 2 m

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 25 kN·m at x = 1 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 5 mM₀ = 25 kN·m clockwise, at 1 m

M_x = kN·m

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment M_max in the beam.

L = 6 mP = 30 kN at 4 mw = 2 kN/m from 0 to 6 m

M_max = kN·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 = 260 mmt_f = 20 mmt_w = 8 mmh_w = 100 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 240 mm plate has a rectangular cut-out 50 mm wide and 50 mm high, centred on the vertical centre line with its centre 96 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 240 mmcut-out: 50 × 50 mmhole centre: y = 96 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 100 MPa, σᵧ = 120 MPa and τₓᵧ = 40 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 100 MPaσ_y = 120 MPaτ_xy = 40 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries an axial tensile force of 190 kN together with a torque of 1500 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 40 mmF = 190 kNT = 1500 N·m

σ₁ = 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 = 12 kN/mL = 5 mE = 70 GPa (aluminium)I = 23 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 8 mw = 8 kN/m

R_B = kN

13.Column bucklingMedium

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

L = 5 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

14.Column bucklingHard

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

L = 1 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 1935 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 = 95·1000/490.9 = 193.5 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 193.5/100000 = 0.001935 = 1935 µε

Answer

ε = 1935 µε

2.Axial stress & strain

Answer: δ = 1.167 mm

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·12²/4 = 113.1 mm²

σ₁ = 28.29 MPa

σ₂ = 176.8 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 28.29·2000/200000 = 0.2829 mm

δ₂ = σ₂L₂/E₂ = 176.8·1000/200000 = 0.8842 mm

δ = δ₁ + δ₂ = 1.167 mm

Answer

δ = 1.167 mm

3.Torsion of shafts

Answer: φ = 4.444 °

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 second moment of area

J = π(30⁴ − 21⁴)/32 = 60428 mm⁴

Angle of twist

T = 250 N·m = 250000 N·mm

L = 1.5 m = 1500 mm

G = 80 GPa = 80000 MPa

φ = TL/(GJ) = 250000·1500/(80000·60428) = 0.07757 rad

φ = 0.07757·180/π = 4.444°

Answer

φ = 4.444°

4.Torsion of shafts

Answer: τ_max = 56.59 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 200 N·m

T_AB = T_C + T_B = 200 + 100 = 300 N·m

Maximum shear stress

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

|T|_max = 300 N·m

τ_max = 16·300000/(π·30³) = 56.59 MPa

Answer

τ_max = 56.59 MPa

5.Shear force & bending moment

Answer: M_x = −5 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 = 25/5 = 5 kN (upward)

R_A = −R_B = −5 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −5·1 = −5 kN·m

Answer

M_x = −5 kN·m

6.Shear force & bending moment

Answer: M_max = 48 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 = 2·6 = 12 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 30·4 + 2·6·3

R_B = 156/6 = 26 kN

ΣF_y = 0: R_A = 42 − 26 = 16 kN

Where is V = 0?

V(4⁻) = R_A − w·4 = 8 kN

V(4⁺) = 8 − 30 = −22 kN

The shear changes sign at the load, so the maximum is under the load.

Maximum moment

M_max = R_A·4 − w·4²/2 = 16·4 − 2·4²/2 = 48 kN·m

Answer

M_max = 48 kN·m

It occurs at x = 4 m from A.

7.Centroid & second moment of area

Answer: I_x = 3.336 × 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 = 260·20 = 5200 mm², y_f = 100 + 20/2 = 110 mm

web: A_w = 8·100 = 800 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (5200·110 + 800·50)/6000 = 102 mm

Parallel-axis theorem

flange: 260·20³/12 + 5200·(110 − 102)² = 506133 mm⁴

web: 8·100³/12 + 800·(50 − 102)² = 2.83 × 10⁶ mm⁴

I_x = 506133 + 2.83 × 10⁶ = 3.336 × 10⁶ mm⁴ = 3.336 × 10⁶ mm⁴

Answer

I_x = 3.336 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 228.4 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (48000·120 − 2500·96)/(48000 − 2500) = 121.3 mm

Full plate about ȳ

200·240³/12 + 48000·−1.319² = 230.5 × 10⁶ mm⁴

Hole about ȳ

50·50³/12 + 2500·−25.32² = 2.123 × 10⁶ mm⁴

Subtract

I_x = 228.4 × 10⁶ mm⁴

Answer

I_x = 228.4 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 41.23 MPa

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.

Radius of Mohr’s circle

τ_max = R = √(((100 − 120)/2)² + 40²) = 41.23 MPa

Answer

τ_max = 41.23 MPa

10.Mohr’s circle & stress transformation

Answer: σ₁ = 216.9 MPa

Stresses at the surface

σ = F/A = 190000/(π·40²/4) = 190000/1257 = 151.2 MPa

τ = 16T/(πd³) = 16·1500000/(π·40³) = 119.4 MPa

Mohr’s circle

C = σ/2 = 75.6 MPa

R = √(75.6² + 119.4²) = 141.3 MPa

σ₁ = C + R = 216.9 MPa

σ₂ = C − R = −65.69 MPa

Answer

σ₁ = 216.9 MPa

11.Beam deflection

Answer: δ_max = 60.66 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 = 12 N/mm, L = 5000 mm

EI = 1.61 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·12·5000⁴/(384·1.61 × 10¹²) = 60.66 mm

Answer

δ_max = 60.66 mm

12.Beam deflection

Answer: R_B = 24 kN

Remove the prop

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

δ_load = wL⁴/(8EI)

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) = wL⁴/(8EI) → R_B = 3wL/8 = 3·8·8/8 = 24 kN

Answer

R_B = 24 kN

13.Column buckling

Answer: P_cr = 102.5 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/3500² = 102511 N = 102.5 kN

Answer

P_cr = 102.5 kN

14.Column buckling

Answer: σ_cr = 111 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 1·1000 = 1000 mm

KL/r = 1000/7.5 = 133.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/133.3² = 111 MPa

Answer

σ_cr = 111 MPa

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