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 400 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 50 mm, length 300 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 300 mm, E = 70 GPaδ = 0.5 mm

P = kN

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of copper (diameter 50 mm, length 400 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: copper, Ø50 mm, 400 mm, E = 120 GPaP = 60 kN

σ₂ = 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 = 100 N·md = 30 mmL = 1.2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 50 mm is fixed at A. A torque of 2000 N·m acts at B, 3 m from A, and a torque of 1500 N·m in the opposite direction acts at the free end C, a further 2 m along. Find the maximum shear stress τ_max in the shaft.

d = 50 mmT_B = 2000 N·mT_C = 1500 N·mAB = 3 mBC = 2 m

τ_max = MPa

5.Shear force & bending momentMedium

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 1 m from A. Find the bending moment M_x at the section x = 0.5 m from the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 15 kN at 1 mw = 8 kN/m from 0 to 4 mx = 0.5 m

M_x = kN·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.

45 kN @ 6 m50 kN @ 2 m8 kN/m, 0–5 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 200 mm wide with two upright legs, 100 mm high overall. Base and legs are all 10 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 200 mmh = 100 mmt = 10 mm

ȳ = 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 size of the bending stress σ_top at the top edge.

b_f = 200 mmt_f = 20 mmt_w = 14 mmh_w = 100 mmM = 10 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 60 MPa, σᵧ = 70 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 60 MPaσ_y = 70 MPaτ_xy = −50 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 50 mm carries a bending moment of 1000 N·m together with a torque of 1000 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τ_max.

d = 50 mmM = 1000 N·mT = 1000 N·m

τ_max = 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 = 8 kN/mL = 3 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ_max = 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 = 5 kN/mL = 4 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 3 mE = 200 GPa (steel)b × h = 30 × 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 110 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 110 kNn = 2L = 5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 132.5 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 = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(70000·1963) = 2.183 × 10⁻⁶ mm/N

Load

P = 0.5/3.774 × 10⁻⁶ = 132477 N = 132.5 kN

Answer

P = 132.5 kN

2.Axial stress & strain

Answer: σ₂ = 17.86 MPa

Stiffnesses

k₁ = 200000·1257/600 = 418879 N/mm

k₂ = 120000·1963/400 = 589049 N/mm

Share of segment 2

F₂ = 60000·589049/1.008 × 10⁶ = 35065 N

Stress

σ₂ = 35065/1963 = 17.86 MPa (compression)

Answer

σ₂ = 17.86 MPa

3.Torsion of shafts

Answer: φ = 1.081 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1200/(80000·79522) = 0.01886 rad

φ = 0.01886·180/π = 1.081°

Answer

φ = 1.081°

4.Torsion of shafts

Answer: τ_max = 61.12 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 1500 N·m

T_AB = T_C − T_B = 1500 − 2000 = −500 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 1500 N·m

τ_max = 16·1.5 × 10⁶/(π·50³) = 61.12 MPa

Answer

τ_max = 61.12 MPa

5.Shear force & bending moment

Answer: M_x = −56.5 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.

Keep the free-end part

P: −15·0.5 = −7.5 kN·m

w: −8·3.5²/2 = −49 kN·m

M_x = −7.5 − 49 = −56.5 kN·m

Answer

M_x = −56.5 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: C (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 = 41 kN, R₂ = 94 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 = 37, M_C = 66 kN·m, M_D = 37, M_E = 0, M_F = −45, M_G = 0

Answer

Mark C: M_C = 66 kN·m

7.Centroid & second moment of area

Answer: ȳ = 28.68 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 = 200·10 = 2000 mm², y₁ = 5 mm

two legs: A₂ = 2·10·90 = 1800 mm², y₂ = 10 + 90/2 = 55 mm

Centroid

ȳ = (2000·5 + 1800·55)/3800 = 28.68 mm

Answer

ȳ = 28.68 mm

8.Centroid & second moment of area

Answer: σ_top = 50.77 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 200·20 = 4000 mm², y_f = 100 + 20/2 = 110 mm

web: A_w = 14·100 = 1400 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4000·110 + 1400·50)/5400 = 94.44 mm

Parallel-axis theorem

flange: 200·20³/12 + 4000·(110 − 94.44)² = 1.101 × 10⁶ mm⁴

web: 14·100³/12 + 1400·(50 − 94.44)² = 3.932 × 10⁶ mm⁴

I_x = 1.101 × 10⁶ + 3.932 × 10⁶ = 5.033 × 10⁶ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 120 − 94.44 = 25.56 mm

σ = M·y/I = 1 × 10⁷·25.56/5.033 × 10⁶ = 50.77 MPa

Answer

σ_top = 50.77 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 50.25 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 = √(((60 − 70)/2)² + −50²) = 50.25 MPa

Answer

τ_max = 50.25 MPa

10.Mohr’s circle & stress transformation

Answer: τ_max = 57.62 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·1000000/(π·50³) = 81.49 MPa

τ = 16T/(πd³) = 16·1000000/(π·50³) = 40.74 MPa

Mohr’s circle

C = σ/2 = 40.74 MPa

R = √(40.74² + 40.74²) = 57.62 MPa

Answer

τ_max = 57.62 MPa

11.Beam deflection

Answer: δ_max = 3.516 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 = 8 N/mm, L = 3000 mm

EI = 2.4 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·3000⁴/(384·2.4 × 10¹²) = 3.516 mm

Answer

δ_max = 3.516 mm

12.Beam deflection

Answer: I_req = 21.43 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·5·4000⁴/(384·70000·11.11) = 2.143 × 10⁷ mm⁴

I_req = 21.43 × 10⁶ mm⁴

Answer

I_req = 21.43 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 24.67 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 = 50·30³/12 = 112500 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·112500/3000² = 24674 N = 24.67 kN

Answer

P_cr = 24.67 kN

14.Column buckling

Answer: d = 94.42 mm

Required critical load

P_cr = n·P = 2·110000 = 220000 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·220000·3500²/(π³·70000))^(1/4) = 94.42 mm

Answer

d = 94.42 mm

Check: the slenderness KL/r is 148.3, so the column is slender and Euler applies.

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