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 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 40 mm, length 200 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, Ø40 mm, 500 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 200 mm, E = 110 GPaδ = 0.25 mm

P = kN

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 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 = 160 MPatube: Ø80/Ø50 mm, E = 100 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 40 N·m. The two ends of a gauge length of 250 mm rotate 2.96° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 40 N·md = 15 mmL = 250 mmφ = 2.96°

G = GPa

4.Torsion of shaftsHard

A hollow shaft must transmit 100 kW at 300 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

P = 100 kWN = 300 rpmd_i/d_o = 0.75τ_allow = 40 MPa

d_o = mm

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 5 kN/m from 0 to 10 mx = 1.5 m

V_x = kN

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.

55 kN @ 0 m25 kN @ 12 m25 kN @ 8 m15 kN/m, 2–10 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 = 240 mmt_f = 15 mmt_w = 8 mmh_w = 260 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 200 mm plate has a rectangular cut-out 60 mm wide and 30 mm high, centred on the vertical centre line with its centre 130 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 200 mmcut-out: 60 × 30 mmhole centre: y = 130 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 60 MPa and σᵧ = 20 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 85 MPa?

σ_x = 60 MPaσ_y = 20 MPaσ₁ = 85 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 130 MPa, σᵧ = −30 MPa and τₓᵧ = −40 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.

σ_x = 130 MPaσ_y = −30 MPaτ_xy = −40 MPaθ = 60°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_P under the load.

P = 7 kNa = 1 mL = 2.5 mE = 70 GPa (aluminium)I = 8.5 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 6 kN/mL = 2 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

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 = 3 mE = 70 GPa (aluminium)b × h = 40 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 135 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 135 kNn = 2L = 3 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 72.75 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 = 1257 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 200/(110000·1257) = 1.447 × 10⁻⁶ mm/N

Load

P = 0.25/3.436 × 10⁻⁶ = 72753 N = 72.75 kN

Answer

P = 72.75 kN

2.Axial stress & strain

Answer: P_max = 268.6 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 100·10³·3063 = 4.477 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0006) = 0.0006 (brass)

Largest load

P_max = 0.0006·4.477 × 10⁸ = 268606 N = 268.6 kN

Answer

P_max = 268.6 kN

3.Torsion of shafts

Answer: G = 38.95 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 = 2.96·π/180 = 0.05166 rad

Shear modulus

T = 40 N·m = 40000 N·mm

G = TL/(Jφ) = 40000·250/(4970·0.05166) = 38946 MPa

G = 38.95 GPa

Answer

G = 38.95 GPa

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

4.Torsion of shafts

Answer: d_o = 84.01 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 100000/31.42 = 3183 N·m

Outer diameter from the allowable stress

T = 3183 N·m = 3.183 × 10⁶ N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·3.183 × 10⁶/(π·40·0.6836)) = 84.01 mm

d_i = 0.75·84.01 = 63.01 mm

Answer

d_o = 84.01 mm

A solid shaft would need d = 74 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: V_x = 17.5 kN

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 = 5·10 = 50 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 5·10·5

R_B = 250/10 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 1.5 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.

V = R_A − (all downward loads left of the cut)

V_x = 25 − 5·1.5 = 17.5 kN

Answer

V_x = 17.5 kN

6.Shear force & bending moment

Answer: B (x = 2 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 = 128.8 kN, R₂ = 96.25 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 = −110 kN·m, M_C = 7.5, M_D = 65, M_E = 62.5, M_F = −50, M_G = 0

Answer

Mark B: M_B = −110 kN·m

7.Centroid & second moment of area

Answer: I_x = 36.71 × 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 = 240·15 = 3600 mm², y_f = 260 + 15/2 = 267.5 mm

web: A_w = 8·260 = 2080 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3600·267.5 + 2080·130)/5680 = 217.1 mm

Parallel-axis theorem

flange: 240·15³/12 + 3600·(267.5 − 217.1)² = 9.195 × 10⁶ mm⁴

web: 8·260³/12 + 2080·(130 − 217.1)² = 2.751 × 10⁷ mm⁴

I_x = 9.195 × 10⁶ + 2.751 × 10⁷ = 3.671 × 10⁷ mm⁴ = 36.71 × 10⁶ mm⁴

Answer

I_x = 36.71 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 131.5 × 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 = (40000·100 − 1800·130)/(40000 − 1800) = 98.59 mm

Full plate about ȳ

200·200³/12 + 40000·1.414² = 133.4 × 10⁶ mm⁴

Hole about ȳ

60·30³/12 + 1800·31.41² = 1.911 × 10⁶ mm⁴

Subtract

I_x = 131.5 × 10⁶ mm⁴

Answer

I_x = 131.5 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 40.31 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.

Required radius

C = (60 + 20)/2 = 40 MPa

R = σ₁ − C = 85 − 40 = 45 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(45² − 20²) = 40.31 MPa

Answer

τ_xy = 40.31 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −24.64 MPa

Transformation equation

σ_θ = 50 + 80·−0.5 + −40·0.866 = −24.64 MPa

Answer

σ_θ = −24.64 MPa

11.Beam deflection

Answer: δ_P = 3.922 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 = 7000 N, a = 1000 mm, L = 2500 mm

EI = 5.95 × 10¹¹ N·mm²

Deflection under the load

δ_P = Pa³/(3EI) = 7000·1000³/(3·5.95 × 10¹¹) = 3.922 mm

Answer

δ_P = 3.922 mm

12.Beam deflection

Answer: I_req = 1.125 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 2000/360 = 5.556 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·6·2000⁴/(384·200000·5.556) = 1.125 × 10⁶ mm⁴

I_req = 1.125 × 10⁶ mm⁴

Answer

I_req = 1.125 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 98.26 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 = 60·40³/12 = 320000 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·320000/1500² = 98257 N = 98.26 kN

Answer

P_cr = 98.26 kN

14.Column buckling

Answer: d = 76.98 mm

Required critical load

P_cr = n·P = 2·135000 = 270000 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·270000·2100²/(π³·70000))^(1/4) = 76.98 mm

Answer

d = 76.98 mm

Check: the slenderness KL/r is 109.1, so the column is slender and Euler applies.

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