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 aluminium bar under the axial load.

P = 50 kNd = 30 mmL = 2.5 mE = 70 GPa

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 115 kNsegment 1: Ø25 mm, 1 m, copper (E = 117 GPa)segment 2: Ø12 mm, 2 m, steel (E = 200 GPa)

σ_max = MPa

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 400 mm rotate 2.31° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 40 N·md = 15 mmL = 400 mmφ = 2.31°

G = GPa

4.Torsion of shaftsHard

A solid brass shaft transmits 5 kW at 1000 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 5 kWN = 1000 rpmτ_allow = 60 MPaθ_allow = 1°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 4 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 2 kN/m from 0 to 8 mx = 4 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 15 kN at 7 mw = 4 kN/m from 0 to 8 mx = 7 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 120 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 52.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 150 mmd = 30 mmhole centre: y = 52.5 mm

I_x = × 10⁶ mm⁴

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 5 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 200 mmt_f = 20 mmt_w = 20 mmh_w = 180 mmM = 5 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = −50 MPa and τₓᵧ = 40 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = −70 MPaσ_y = −50 MPaτ_xy = 40 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 140 MPa, σᵧ = 90 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σ_x = 140 MPaσ_y = 90 MPaτ_xy = 30 MPaθ = 15°

σ_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 = 10 kN/mL = 6 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 5 mw = 15 kN/m

R_B = kN

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 = 6 mE = 70 GPa (aluminium)b × h = 20 × 80 mm

P_cr = kN

14.Column bucklingHard

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

L = 3 mE = 200 GPa (steel)d = 25 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 2.526 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·30²/4 = 706.9 mm²

σ = P/A = 50·1000/706.9 = 70.74 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 70.74/70000 = 0.001011 = 1011 µε

Elongation

δ = ε·L = 0.001011·2500 mm = 2.526 mm

Answer

δ = 2.526 mm

2.Axial stress & strain

Answer: σ_max = 1017 MPa

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 = 234.3 MPa

σ₂ = 1017 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 1017 MPa

Answer

σ_max = 1017 MPa

3.Torsion of shafts

Answer: G = 79.85 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 = 2.31·π/180 = 0.04032 rad

Shear modulus

T = 40 N·m = 40000 N·mm

G = TL/(Jφ) = 40000·400/(4970·0.04032) = 79848 MPa

G = 79.85 GPa

Answer

G = 79.85 GPa

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

4.Torsion of shafts

Answer: d = 29.07 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 5000/104.7 = 47.75 N·m

Diameter for strength

T = 47.75 N·m = 47746 N·mm

d_τ = ∛(16·47746/(π·60)) = 15.94 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·47746/(π·39000·1.745 × 10⁻⁵))^1/4 = 29.07 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(15.94, 29.07) = 29.07 mm

Answer

d = 29.07 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = 16 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 = 2·8 = 16 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 2·8·4

R_B = 64/8 = 8 kN

ΣF_y = 0: R_A = 16 − 8 = 8 kN

Cut at x = 4 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 = 8·4 − 2·4·2 = 16 kN·m

Answer

M_x = 16 kN·m

6.Shear force & bending moment

Answer: M_x = 27.13 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 4·8 = 32 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 15·7 + 4·8·4

R_B = 233/8 = 29.13 kN

ΣF_y = 0: R_A = 47 − 29.13 = 17.88 kN

Cut at x = 7 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 = 17.88·7 − 4·7·3.5 = 27.13 kN·m

Answer

M_x = 27.13 kN·m

7.Centroid & second moment of area

Answer: I_x = 33.34 × 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 = (18000·75 − 706.9·52.5)/(18000 − 706.9) = 75.92 mm

Full plate about ȳ

120·150³/12 + 18000·−0.9197² = 33.77 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−23.42² = 0.4275 × 10⁶ mm⁴

Subtract

I_x = 33.34 × 10⁶ mm⁴

Answer

I_x = 33.34 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 9.96 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 = 200·20 = 4000 mm², y_f = 180 + 20/2 = 190 mm

web: A_w = 20·180 = 3600 mm², y_w = 180/2 = 90 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4000·190 + 3600·90)/7600 = 142.6 mm

Parallel-axis theorem

flange: 200·20³/12 + 4000·(190 − 142.6)² = 9.108 × 10⁶ mm⁴

web: 20·180³/12 + 3600·(90 − 142.6)² = 1.969 × 10⁷ mm⁴

I_x = 9.108 × 10⁶ + 1.969 × 10⁷ = 2.88 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 200 − 142.6 = 57.37 mm

σ = M·y/I = 5 × 10⁶·57.37/2.88 × 10⁷ = 9.96 MPa

Answer

σ_top = 9.96 MPa

9.Mohr’s circle & stress transformation

Answer: τ_abs = 50.62 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 = −60 MPa, R = √((−10)² + 40²) = 41.23 MPa

σ₁ = −18.77 MPa, σ₂ = −101.2 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(82.46, 18.77, 101.2)/2 = 50.62 MPa

Answer

τ_abs = 50.62 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 = 151.7 MPa

Transformation equation

σ_θ = 115 + 25·0.866 + 30·0.5 = 151.7 MPa

Answer

σ_θ = 151.7 MPa

11.Beam deflection

Answer: δ_max = 28.7 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 = 6000 mm

EI = 5.88 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·10·6000⁴/(384·5.88 × 10¹²) = 28.7 mm

Answer

δ_max = 28.7 mm

12.Beam deflection

Answer: R_B = 28.13 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·15·5/8 = 28.13 kN

Answer

R_B = 28.13 kN

13.Column buckling

Answer: P_cr = 4.094 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 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·53333/3000² = 4094 N = 4.094 kN

Answer

P_cr = 4.094 kN

14.Column buckling

Answer: σ_cr = 17.48 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 0.7·3000 = 2100 mm

KL/r = 2100/6.25 = 336

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/336² = 17.48 MPa

Answer

σ_cr = 17.48 MPa

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