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 steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 50 mm, E = 70 GPa). Both are 500 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 70 GPaL = 500 mm

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 = 90 kNsegment 1: Ø35 mm, 2 m, copper (E = 117 GPa)segment 2: Ø20 mm, 0.5 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid steel shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 1000 N·md = 50 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 1450 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 20 kWN = 1450 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 8.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 3 kN/m from 0 to 10 mx = 8.5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 25 kN at 1 mw = 6 kN/m from 0 to 8 mx = 1 m

M_x = 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 = 240 mmt_f = 25 mmt_w = 12 mmh_w = 100 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 40 MPa in tension and 90 MPa in compression. Find the largest hogging bending moment M_allow the beam can carry (give its size).

b_f = 180 mmt_f = 25 mmt_w = 14 mmh_w = 260 mmσ_t,allow = 40 MPaσ_c,allow = 90 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = 100 MPa and τₓᵧ = 10 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 80 MPaσ_y = 100 MPaτ_xy = 10 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 110 MPa, σᵧ = 0 MPa and τₓᵧ = 20 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 35° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 110 MPaσ_y = 0 MPaτ_xy = 20 MPaθ = 35°

τ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 6 kN/mL = 3 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported beam carries two equal point loads P at its third points (L/3 from each support). Using superposition, find the midspan deflection δ_max.

P = 26 kNL = 6 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴

δ_max = 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 = 6 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 110 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 110 kNn = 3L = 5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.1469 mm

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

A_r = π·25²/4 = 490.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·5890 = 5.105 × 10⁸ N

Shortening

δ = 150000·500/5.105 × 10⁸ = 0.1469 mm

Answer

δ = 0.1469 mm

2.Axial stress & strain

Answer: σ_max = 286.5 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·35²/4 = 962.1 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 93.54 MPa

σ₂ = 286.5 MPa

Largest stress

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

σ_max = 286.5 MPa

Answer

σ_max = 286.5 MPa

3.Torsion of shafts

Answer: φ = 1.751 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 1000·1000·1500/(80000·613592) = 0.03056 rad

φ = 0.03056·180/π = 1.751°

Answer

φ = 1.751°

4.Torsion of shafts

Answer: d = 25.6 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 20000/151.8 = 131.7 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·131714/(π·40)) = 25.6 mm

Answer

d = 25.6 mm

In practice you would round up to the next standard size.

5.Shear force & bending moment

Answer: M_x = 19.13 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·10 = 30 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 3·10·5

R_B = 150/10 = 15 kN

ΣF_y = 0: R_A = 30 − 15 = 15 kN

Cut at x = 8.5 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 15·8.5 − 3·8.5·4.25 = 19.13 kN·m

Answer

M_x = 19.13 kN·m

6.Shear force & bending moment

Answer: M_x = 42.88 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·8 = 48 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 25·1 + 6·8·4

R_B = 217/8 = 27.13 kN

ΣF_y = 0: R_A = 73 − 27.13 = 45.88 kN

Cut at x = 1 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 45.88·1 − 6·1·0.5 = 42.88 kN·m

Answer

M_x = 42.88 kN·m

7.Centroid & second moment of area

Answer: I_x = 5.219 × 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 = 240·25 = 6000 mm², y_f = 100 + 25/2 = 112.5 mm

web: A_w = 12·100 = 1200 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·112.5 + 1200·50)/7200 = 102.1 mm

Parallel-axis theorem

flange: 240·25³/12 + 6000·(112.5 − 102.1)² = 963542 mm⁴

web: 12·100³/12 + 1200·(50 − 102.1)² = 4.255 × 10⁶ mm⁴

I_x = 963542 + 4.255 × 10⁶ = 5.219 × 10⁶ mm⁴ = 5.219 × 10⁶ mm⁴

Answer

I_x = 5.219 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 26.56 kN·m

Centroid

ȳ = (4500·272.5 + 3640·130)/8140 = 208.8 mm

top edge: c = 285 − 208.8 = 76.22 mm

Second moment of area

I = 180·25³/12 + 4500·63.72² + 14·260³/12 + 3640·78.78² = 6.16 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 40·6.16 × 10⁷/76.22 = 32.33 kN·m

bottom edge: M = 90·6.16 × 10⁷/208.8 = 26.56 kN·m

Governing value

M_allow = min = 26.56 kN·m (bottom edge)

Answer

M_allow = 26.56 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_abs = 52.07 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 90 MPa, R = √((−10)² + 10²) = 14.14 MPa

σ₁ = 104.1 MPa, σ₂ = 75.86 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(28.28, 104.1, 75.86)/2 = 52.07 MPa

Answer

τ_abs = 52.07 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: τ_θ = −44.84 MPa

Transformation equation

τ_θ = −55·0.9397 + 20·0.342 = −44.84 MPa

Answer

τ_θ = −44.84 MPa

11.Beam deflection

Answer: δ = 13.21 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 = 6 N/mm, L = 3000 mm

EI = 4.6 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 6·3000⁴/(8·4.6 × 10¹²) = 13.21 mm

Answer

δ = 13.21 mm

12.Beam deflection

Answer: δ_max = 27.69 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 = 26000 N, L = 6000 mm

EI = 7.2 × 10¹² N·mm²

One load at L/3

δ₁ = Pb(3L² − 4b²)/(48EI), b = L/3 = 2000 mm

δ₁ = 13.84 mm

Add both loads

δ_max = 2·δ₁ = 23PL³/(648EI) = 27.69 mm

Answer

δ_max = 27.69 mm

13.Column buckling

Answer: P_cr = 38.59 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/6000² = 38586 N = 38.59 kN

Answer

P_cr = 38.59 kN

14.Column buckling

Answer: d = 67.92 mm

Required critical load

P_cr = n·P = 3·110000 = 330000 N

KL = 0.5·5000 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·330000·2500²/(π³·200000))^(1/4) = 67.92 mm

Answer

d = 67.92 mm

Check: the slenderness KL/r is 147.2, so the column is slender and Euler applies.

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