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 30 mm, length 300 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 0.2 mm in total.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 200 mm, E = 110 GPaδ = 0.2 mm

P = kN

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 50 mm, E = 70 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 = 120 MPatube: Ø100/Ø50 mm, E = 70 GPa, σ_allow = 60 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid brass 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 = 40 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 300 rpmd_o = 75 mmd_i = 38 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.

45 kN @ 2 m40 kN @ 4 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.

15 kN @ 0 m30 kN @ 6 m10 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 160 mm wide with two upright legs, 170 mm high overall. Base and legs are all 8 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 160 mmh = 170 mmt = 8 mm

ȳ = mm

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 50 MPa in tension and 120 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 200 mmt_f = 15 mmt_w = 10 mmh_w = 160 mmσ_t,allow = 50 MPaσ_c,allow = 120 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 = −60 MPa and τₓᵧ = −60 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 = −60 MPaτ_xy = −60 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 60 MPa, σᵧ = 100 MPa and τₓᵧ = 50 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 = 60 MPaσ_y = 100 MPaτ_xy = 50 MPaθ = 60°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.

P = 13 kNw = 2 kN/mL = 2 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴

δ = 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 = 15 kN/mL = 4 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 4 mE = 200 GPa (steel)d = 50 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 60 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 60 kNn = 2L = 2 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 56.04 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 = 706.9 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N

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

Load

P = 0.2/3.569 × 10⁻⁶ = 56039 N = 56.04 kN

Answer

P = 56.04 kN

2.Axial stress & strain

Answer: P_max = 332.2 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·5890 = 5.537 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0008571) = 0.0006 (steel)

Largest load

P_max = 0.0006·5.537 × 10⁸ = 332223 N = 332.2 kN

Answer

P_max = 332.2 kN

3.Torsion of shafts

Answer: φ = 5.845 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 1000·1000·1000/(39000·251327) = 0.102 rad

φ = 0.102·180/π = 5.845°

Answer

φ = 5.845°

4.Torsion of shafts

Answer: τ_max = 8.228 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 20000/31.42 = 636.6 N·m

Polar second moment of area

J = π(75⁴ − 38⁴)/32 = 2.902 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 636620·37.5/2.902 × 10⁶ = 8.228 MPa

Answer

τ_max = 8.228 MPa

5.Shear force & bending moment

Answer: C (x = 4 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 = 64.17 kN, R₂ = 20.83 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 = 128.3, M_C = 166.7 kN·m, M_D = 125, M_E = 83.33, M_F = 41.67, M_G = 0

Answer

Mark C: M_C = 166.7 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 = 15 kN, R₂ = 60 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 = −15, M_C = −20, M_D = −35, M_E = −60 kN·m, M_F = −30, M_G = 0

Answer

Mark E: M_E = −60 kN·m

7.Centroid & second moment of area

Answer: ȳ = 60.9 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 = 160·8 = 1280 mm², y₁ = 4 mm

two legs: A₂ = 2·8·162 = 2592 mm², y₂ = 8 + 162/2 = 89 mm

Centroid

ȳ = (1280·4 + 2592·89)/3872 = 60.9 mm

Answer

ȳ = 60.9 mm

8.Centroid & second moment of area

Answer: M_allow = 4.18 kN·m

Centroid

ȳ = (3000·167.5 + 1600·80)/4600 = 137.1 mm

top edge: c = 175 − 137.1 = 37.93 mm

Second moment of area

I = 200·15³/12 + 3000·30.43² + 10·160³/12 + 1600·57.07² = 1.146 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 120·1.146 × 10⁷/37.93 = 36.25 kN·m

bottom edge: M = 50·1.146 × 10⁷/137.1 = 4.18 kN·m

Governing value

M_allow = min = 4.18 kN·m (bottom edge)

Answer

M_allow = 4.18 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_abs = 65.41 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 = −70 MPa, R = √((−10)² + (−60)²) = 60.83 MPa

σ₁ = −9.172 MPa, σ₂ = −130.8 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(121.7, 9.172, 130.8)/2 = 65.41 MPa

Answer

τ_abs = 65.41 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: σ_θ = 133.3 MPa

Transformation equation

σ_θ = 80 + −20·−0.5 + 50·0.866 = 133.3 MPa

Answer

σ_θ = 133.3 MPa

11.Beam deflection

Answer: δ = 41.85 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 = 13000 N, w = 2 N/mm, L = 2000 mm

EI = 9.24 × 10¹¹ N·mm²

Each load on its own

δ_P = PL³/(3EI) = 13000·2000³/(3·9.24 × 10¹¹) = 37.52 mm

δ_w = wL⁴/(8EI) = 2·2000⁴/(8·9.24 × 10¹¹) = 4.329 mm

Add them

δ = 37.52 + 4.329 = 41.85 mm

Answer

δ = 41.85 mm

12.Beam deflection

Answer: I_req = 64.29 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·15·4000⁴/(384·70000·11.11) = 6.429 × 10⁷ mm⁴

I_req = 64.29 × 10⁶ mm⁴

Answer

I_req = 64.29 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 9.462 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/8000² = 9462 N = 9.462 kN

Answer

P_cr = 9.462 kN

14.Column buckling

Answer: d = 51.32 mm

Required critical load

P_cr = n·P = 2·60000 = 120000 N

KL = 0.7·2000 = 1400 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·120000·1400²/(π³·70000))^(1/4) = 51.32 mm

Answer

d = 51.32 mm

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

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