Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 aluminium bar, in microstrain (µε = 10⁻⁶).

P = 25 kNd = 20 mmL = 1.5 mE = 70 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 = 60 kNsegment 1: Ø40 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1.5 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 50 mm and length 2 m; segment BC has diameter 25 mm and length 2.5 m. A torque of 250 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 250 N·md₁ = 50 mm, L₁ = 2 md₂ = 25 mm, L₂ = 2.5 m

τ_max = MPa

4.Torsion of shaftsHard

A hollow shaft transmits 5 kW at 500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 5 kWN = 500 rpmd_o = 50 mmd_i = 25 mm

τ_max = MPa

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 6.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 12 kN/m from 0 to 8 mx = 6.5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 40 kN at 6 mw = 12 kN/m from 0 to 8 mx = 7.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A hollow rectangular (box) section has outer width B = 220 mm, outer height H = 350 mm and a uniform wall thickness t = 14 mm. Find its radius of gyration r about the horizontal centroidal axis.

B = 220 mmH = 350 mmt = 14 mm

r = mm

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 10 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 280 mmt_f = 10 mmt_w = 12 mmh_w = 200 mmM = 10 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 140 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 140 MPaσ_y = 50 MPaτ_xy = 30 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 80 MPa, σᵧ = 100 MPa and τₓᵧ = 20 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 80 MPaσ_y = 100 MPaτ_xy = 20 MPaθ = 15°

τ_θ = 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 = 3 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴

δ_max = 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 = 11 kNw = 2 kN/mL = 2 mE = 70 GPa (aluminium)I = 8.5 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 6 mE = 70 GPa (aluminium)b × h = 50 × 100 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 4 mE = 200 GPa (steel)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 1137 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·20²/4 = 314.2 mm²

σ = P/A = 25·1000/314.2 = 79.58 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 79.58/70000 = 0.001137 = 1137 µε

Answer

ε = 1137 µε

2.Axial stress & strain

Answer: δ = 4.331 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·40²/4 = 1257 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 47.75 MPa

σ₂ = 191 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 47.75·500/100000 = 0.2387 mm

δ₂ = σ₂L₂/E₂ = 191·1500/70000 = 4.093 mm

δ = δ₁ + δ₂ = 4.331 mm

Answer

δ = 4.331 mm

3.Torsion of shafts

Answer: τ_max = 81.49 MPa

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.

Stress in each segment

T = 250 N·m = 250000 N·mm

τ_AB = 16·250000/(π·50³) = 10.19 MPa

τ_BC = 16·250000/(π·25³) = 81.49 MPa

Answer

τ_max = 81.49 MPa

4.Torsion of shafts

Answer: τ_max = 4.15 MPa

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 5000/52.36 = 95.49 N·m

Polar second moment of area

J = π(50⁴ − 25⁴)/32 = 575243 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 95493·25/575243 = 4.15 MPa

Answer

τ_max = 4.15 MPa

5.Shear force & bending moment

Answer: M_x = 58.5 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·8 = 96 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 12·8·4

R_B = 384/8 = 48 kN

ΣF_y = 0: R_A = 96 − 48 = 48 kN

Cut at x = 6.5 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 = 48·6.5 − 12·6.5·3.25 = 58.5 kN·m

Answer

M_x = 58.5 kN·m

6.Shear force & bending moment

Answer: M_x = 37.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·8 = 96 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 40·6 + 12·8·4

R_B = 624/8 = 78 kN

ΣF_y = 0: R_A = 136 − 78 = 58 kN

Cut at x = 7.5 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 = 58·7.5 − 40·(7.5 − 6) − 12·7.5·3.75 = 37.5 kN·m

Answer

M_x = 37.5 kN·m

7.Centroid & second moment of area

Answer: r = 128.8 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

b = 192 mm, h = 322 mm

A = BH − bh = 220·350 − 192·322 = 15176 mm²

I = (BH³ − bh³)/12 = 2.519 × 10⁸ mm⁴

Radius of gyration

r = √(I/A) = √(2.519 × 10⁸/15176) = 128.8 mm

Answer

r = 128.8 mm

8.Centroid & second moment of area

Answer: σ_bottom = 70.29 MPa

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 = 280·10 = 2800 mm², y_f = 200 + 10/2 = 205 mm

web: A_w = 12·200 = 2400 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2800·205 + 2400·100)/5200 = 156.5 mm

Parallel-axis theorem

flange: 280·10³/12 + 2800·(205 − 156.5)² = 6.599 × 10⁶ mm⁴

web: 12·200³/12 + 2400·(100 − 156.5)² = 1.567 × 10⁷ mm⁴

I_x = 6.599 × 10⁶ + 1.567 × 10⁷ = 2.227 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 156.5 mm

σ = M·y/I = 1 × 10⁷·156.5/2.227 × 10⁷ = 70.29 MPa

Answer

σ_bottom = 70.29 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 54.08 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 = √(((140 − 50)/2)² + 30²) = 54.08 MPa

Answer

τ_max = 54.08 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 22.32 MPa

Transformation equation

τ_θ = −−10·0.5 + 20·0.866 = 22.32 MPa

Answer

τ_θ = 22.32 MPa

11.Beam deflection

Answer: δ_max = 20.19 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 = 3000 mm

EI = 6.27 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·12·3000⁴/(384·6.27 × 10¹¹) = 20.19 mm

Answer

δ_max = 20.19 mm

12.Beam deflection

Answer: δ_max = 3.782 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 = 11000 N, w = 2 N/mm, L = 2000 mm, EI = 5.95 × 10¹¹ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 11000·2000³/(48·5.95 × 10¹¹) = 3.081 mm

δ_w = 5wL⁴/(384EI) = 5·2·2000⁴/(384·5.95 × 10¹¹) = 0.7003 mm

Add them

δ_max = 3.081 + 0.7003 = 3.782 mm

Answer

δ_max = 3.782 mm

13.Column buckling

Answer: P_cr = 4.998 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 = 100·50³/12 = 1.042 × 10⁶ mm⁴

Effective length

KL = 2·6000 = 12000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·1.042 × 10⁶/12000² = 4998 N = 4.998 kN

Answer

P_cr = 4.998 kN

14.Column buckling

Answer: σ_cr = 12.34 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 2·4000 = 8000 mm

KL/r = 8000/20 = 400

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/400² = 12.34 MPa

Answer

σ_cr = 12.34 MPa

ExamProblems · examproblems.com/courses/mechanics-of-materials