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 steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa). Both are 800 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.

P = 250 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 120 GPaL = 800 mm

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: Ø35 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 0.5 m, steel (E = 200 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid steel shaft 0.5 m long carries a torque of 2000 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.

T = 2000 N·mL = 0.5 mG = 80 GPaφ_allow = 0.5°

d = mm

4.Torsion of shaftsHard

A hollow shaft must transmit 100 kW at 300 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter d_o.

P = 100 kWN = 300 rpmd_i/d_o = 0.7τ_allow = 60 MPa

d_o = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 25 kN·m at x = 8 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 10 mM₀ = 25 kN·m clockwise, at 8 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 60 kN at 2.5 mw = 3 kN/m from 0 to 4 mx = 3.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 240 mm wide with two upright legs, 90 mm high overall. Base and legs are all 18 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 240 mmh = 90 mmt = 18 mm

ȳ = mm

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 200 mm wide and 22 mm thick, an overall depth of 460 mm and a web 8 mm thick. It carries a bending moment of 200 kN·m about the axis x. Find the size of the bending stress σ_j at the junction of the web and the top flange.

B = 200 mmH = 460 mmt_f = 22 mmt_w = 8 mmM = 200 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 40 MPa, σᵧ = 80 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 = 40 MPaσ_y = 80 MPaτ_xy = 40 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 90 MPa, σᵧ = 40 MPa and τₓᵧ = −60 MPa (σ_z = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 90 MPaσ_y = 40 MPaτ_xy = −60 MPaσ_Y = 355 MPa

n =

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 = 4 mE = 70 GPa (aluminium)I = 120 × 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 = 12 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 = 3 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 130 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 130 kNn = 2.5L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.1929 mm

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

A_r = π·30²/4 = 706.9 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·7461 = 1.037 × 10⁹ N

Shortening

δ = 250000·800/1.037 × 10⁹ = 0.1929 mm

Answer

δ = 0.1929 mm

2.Axial stress & strain

Answer: σ_max = 572 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·35²/4 = 962.1 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 119.5 MPa

σ₂ = 572 MPa

Largest stress

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

σ_max = 572 MPa

Answer

σ_max = 572 MPa

3.Torsion of shafts

Answer: d = 61.8 mm

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.

Required polar moment

T = 2000 N·m = 2 × 10⁶ N·mm

L = 0.5 m = 500 mm

G = 80 GPa = 80000 MPa

φ = 0.5·π/180 = 0.008727 rad

J = TL/(Gφ) = 2 × 10⁶·500/(80000·0.008727) = 1.432 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 61.8 mm

Answer

d = 61.8 mm

In practice you would round up to the next standard size.

4.Torsion of shafts

Answer: d_o = 70.84 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 100000/31.42 = 3183 N·m

Outer diameter from the allowable stress

T = 3183 N·m = 3.183 × 10⁶ N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·3.183 × 10⁶/(π·60·0.7599)) = 70.84 mm

d_i = 0.7·70.84 = 49.59 mm

Answer

d_o = 70.84 mm

A solid shaft would need d = 64.65 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: M_x = −20 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 25/10 = 2.5 kN (upward)

R_A = −R_B = −2.5 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −2.5·8 = −20 kN·m

Answer

M_x = −20 kN·m

6.Shear force & bending moment

Answer: M_x = 21.38 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·4 = 12 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 60·2.5 + 3·4·2

R_B = 174/4 = 43.5 kN

ΣF_y = 0: R_A = 72 − 43.5 = 28.5 kN

Cut at x = 3.5 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 = 28.5·3.5 − 60·(3.5 − 2.5) − 3·3.5·1.75 = 21.38 kN·m

Answer

M_x = 21.38 kN·m

7.Centroid & second moment of area

Answer: ȳ = 25.88 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 = 240·18 = 4320 mm², y₁ = 9 mm

two legs: A₂ = 2·18·72 = 2592 mm², y₂ = 18 + 72/2 = 54 mm

Centroid

ȳ = (4320·9 + 2592·54)/6912 = 25.88 mm

Answer

ȳ = 25.88 mm

8.Centroid & second moment of area

Answer: σ_j = 88.43 MPa

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 460 − 2·22 = 416 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (200·460³ − 192·416³)/12

I_x = 4.704 × 10⁸ mm⁴

Bending stress

y = 230 − 22 = 208 mm

σ = M·y/I = 2 × 10⁸·208/4.704 × 10⁸ = 88.43 MPa

Answer

σ_j = 88.43 MPa

9.Mohr’s circle & stress transformation

Answer: τ_abs = 52.36 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 = √((−20)² + 40²) = 44.72 MPa

σ₁ = 104.7 MPa, σ₂ = 15.28 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(89.44, 104.7, 15.28)/2 = 52.36 MPa

Answer

τ_abs = 52.36 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: n = 2.731

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 65 MPa, R = 65 MPa

σ₁ = 130 MPa, σ₂ = 0 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 130 MPa

n = 355/130 = 2.731

Answer

n = 2.731

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ_max = 3.968 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 = 4000 mm

EI = 8.4 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·10·4000⁴/(384·8.4 × 10¹²) = 3.968 mm

Answer

δ_max = 3.968 mm

12.Beam deflection

Answer: I_req = 51.43 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·12·4000⁴/(384·70000·11.11) = 5.143 × 10⁷ mm⁴

I_req = 51.43 × 10⁶ mm⁴

Answer

I_req = 51.43 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 110.2 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/6000² = 110245 N = 110.2 kN

Answer

P_cr = 110.2 kN

14.Column buckling

Answer: d = 62.02 mm

Required critical load

P_cr = n·P = 2.5·130000 = 325000 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·325000·2100²/(π³·200000))^(1/4) = 62.02 mm

Answer

d = 62.02 mm

Check: the slenderness KL/r is 135.4, so the column is slender and Euler applies.

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