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 elongation δ of the brass bar under the axial load.

P = 80 kNd = 20 mmL = 2.5 mE = 100 GPa

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 25 kNsegment 1: Ø25 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 2 m, steel (E = 200 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid steel shaft is fixed at A. Segment AB has diameter 75 mm and length 1.5 m; segment BC has diameter 40 mm and length 1.5 m. A torque of 1000 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 1000 N·md₁ = 75 mm, L₁ = 1.5 md₂ = 40 mm, L₂ = 1.5 mG = 80 GPa

φ_C = °

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 1000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 1000 rpmd_o = 60 mmd_i = 36 mm

τ_max = MPa

5.Shear force & bending momentMedium

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.

20 kN @ 8 m10 kN/m, 0–8 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)

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.

40 kN @ 0 m35 kN @ 6 m45 kN @ 2 m15 kN/m, 1–4 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 180 mm wide with two upright legs, 80 mm high overall. Base and legs are all 10 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 180 mmh = 80 mmt = 10 mm

ȳ = mm

8.Centroid & second moment of areaHard

A 120 mm × 300 mm plate has a rectangular cut-out 50 mm wide and 40 mm high, centred on the vertical centre line with its centre 105 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 300 mmcut-out: 50 × 40 mmhole centre: y = 105 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 100 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 = 100 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 = 10 MPa, σᵧ = 40 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 20° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 10 MPaσ_y = 40 MPaτ_xy = 30 MPaθ = -20°

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 = 10 kN/mL = 2 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth h_req of the section.

w = 2.5 kN/mL = 3.5 mE = 11 GPa (timber)b = 100 mm

h_req = 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 = 5 mE = 70 GPa (aluminium)b × h = 30 × 80 mm

P_cr = kN

14.Column bucklingHard

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

L = 5 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 6.366 mm

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 = 80·1000/314.2 = 254.6 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 254.6/100000 = 0.002546 = 2546 µε

Elongation

δ = ε·L = 0.002546·2500 mm = 6.366 mm

Answer

δ = 6.366 mm

2.Axial stress & strain

Answer: δ = 2.938 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·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 50.93 MPa

σ₂ = 221 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 50.93·1000/70000 = 0.7276 mm

δ₂ = σ₂L₂/E₂ = 221·2000/200000 = 2.21 mm

δ = δ₁ + δ₂ = 2.938 mm

Answer

δ = 2.938 mm

3.Torsion of shafts

Answer: φ_C = 4.62 °

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 moments

J₁ = π·75⁴/32 = 3.106 × 10⁶ mm⁴

J₂ = π·40⁴/32 = 251327 mm⁴

Twist of each segment

T = 1000 N·m = 1 × 10⁶ N·mm

G = 80 GPa = 80000 MPa

φ_AB = 1 × 10⁶·1500/(80000·3.106 × 10⁶) = 0.006036 rad

φ_BC = 1 × 10⁶·1500/(80000·251327) = 0.0746 rad

Total twist

φ_C = 0.006036 + 0.0746 = 0.08064 rad = 4.62°

Answer

φ_C = 4.62°

4.Torsion of shafts

Answer: τ_max = 5.174 MPa

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 20000/104.7 = 191 N·m

Polar second moment of area

J = π(60⁴ − 36⁴)/32 = 1.107 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 190986·30/1.107 × 10⁶ = 5.174 MPa

Answer

τ_max = 5.174 MPa

5.Shear force & bending moment

Answer: D (x = 6 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 = 60 kN, R₂ = 40 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 = 100, M_C = 160, M_D = 180 kN·m, M_E = 160, M_F = 80, M_G = 0

Answer

Mark D: M_D = 180 kN·m

6.Shear force & bending moment

Answer: E (x = 4 m)

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 82.5 kN, R₂ = 82.5 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 = −40, M_C = −5, M_D = −30, M_E = −70 kN·m, M_F = −35, M_G = 0

Answer

Mark E: M_E = −70 kN·m

7.Centroid & second moment of area

Answer: ȳ = 22.5 mm

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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 180·10 = 1800 mm², y₁ = 5 mm

two legs: A₂ = 2·10·70 = 1400 mm², y₂ = 10 + 70/2 = 45 mm

Centroid

ȳ = (1800·5 + 1400·45)/3200 = 22.5 mm

Answer

ȳ = 22.5 mm

8.Centroid & second moment of area

Answer: I_x = 265.4 × 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 = (36000·150 − 2000·105)/(36000 − 2000) = 152.6 mm

Full plate about ȳ

120·300³/12 + 36000·−2.647² = 270.3 × 10⁶ mm⁴

Hole about ȳ

50·40³/12 + 2000·−47.65² = 4.807 × 10⁶ mm⁴

Subtract

I_x = 265.4 × 10⁶ mm⁴

Answer

I_x = 265.4 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_abs = 55 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 = 100 MPa, R = √(0² + (−10)²) = 10 MPa

σ₁ = 110 MPa, σ₂ = 90 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(20, 110, 90)/2 = 55 MPa

Answer

τ_abs = 55 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 = 13.34 MPa

Transformation equation

τ_θ = −−15·−0.6428 + 30·0.766 = 13.34 MPa

Answer

τ_θ = 13.34 MPa

11.Beam deflection

Answer: δ = 8.333 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 = 10 N/mm, L = 2000 mm

EI = 2.4 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 10·2000⁴/(8·2.4 × 10¹²) = 8.333 mm

Answer

δ = 8.333 mm

12.Beam deflection

Answer: h_req = 176.3 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.

w = 2.5 N/mm, L = 3500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 3500/360 = 9.722 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2.5·3500⁴/(384·11000·9.722) = 4.568 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·4.568 × 10⁷/100)^1/3 = 176.3 mm

Answer

h_req = 176.3 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 4.974 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 1·5000 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·180000/5000² = 4974 N = 4.974 kN

Answer

P_cr = 4.974 kN

14.Column buckling

Answer: σ_cr = 11.05 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 = 0.5·5000 = 2500 mm

KL/r = 2500/10 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/250² = 11.05 MPa

Answer

σ_cr = 11.05 MPa

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