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 40 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa). Both are 800 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø40 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 70 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 total elongation δ.

P = 60 kNsegment 1: Ø30 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid aluminium 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 = 500 N·md = 40 mmL = 2.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 750 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 20 kWN = 750 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 20 kN·m at x = 8 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 10 mM₀ = 20 kN·m clockwise, at 8 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.

60 kN @ 0 m60 kN @ 6 m15 kN @ 2 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

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 260 mmt_f = 15 mmt_w = 8 mmh_w = 180 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 250 mm wide and 18 mm thick, an overall depth of 340 mm and a web 10 mm thick. It carries a bending moment of 80 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 = 250 mmH = 340 mmt_f = 18 mmt_w = 10 mmM = 80 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 110 MPa, σᵧ = 20 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 110 MPaσ_y = 20 MPaτ_xy = 10 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 20 MPa, σᵧ = 90 MPa and τₓᵧ = −30 MPa (σ_z = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 20 MPaσ_y = 90 MPaτ_xy = −30 MPaσ_Y = 460 MPa

n =

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 = 10 kNw = 2 kN/mL = 3 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ = 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 size of the bending moment M_A at the fixed end.

L = 5 mw = 12 kN/m

M_A = kN·m

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 = 1.5 mE = 70 GPa (aluminium)d = 60 mm

P_cr = kN

14.Column bucklingHard

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

L = 3 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.1034 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·40²/4 = 1257 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·7461 = 7.736 × 10⁸ N

Shortening

δ = 100000·800/7.736 × 10⁸ = 0.1034 mm

Answer

δ = 0.1034 mm

2.Axial stress & strain

Answer: δ = 2.873 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·30²/4 = 706.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 84.88 MPa

σ₂ = 530.5 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 84.88·500/70000 = 0.6063 mm

δ₂ = σ₂L₂/E₂ = 530.5·500/117000 = 2.267 mm

δ = δ₁ + δ₂ = 2.873 mm

Answer

δ = 2.873 mm

3.Torsion of shafts

Answer: φ = 10.96 °

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) = 500·1000·2500/(26000·251327) = 0.1913 rad

φ = 0.1913·180/π = 10.96°

Answer

φ = 10.96°

4.Torsion of shafts

Answer: d = 27.86 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 20000/78.54 = 254.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·254648/(π·60)) = 27.86 mm

Answer

d = 27.86 mm

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

5.Shear force & bending moment

Answer: M_x = 4 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 = 20/10 = 2 kN (upward)

R_A = −R_B = −2 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −2·8 + 20 = 4 kN·m

Answer

M_x = 4 kN·m

6.Shear force & bending moment

Answer: E (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 = 50 kN, R₂ = 85 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 = −60, M_C = −70, M_D = −95, M_E = −120 kN·m, M_F = −60, M_G = 0

Answer

Mark E: M_E = −120 kN·m

7.Centroid & second moment of area

Answer: I_x = 13.96 × 10⁶ mm⁴

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 = 260·15 = 3900 mm², y_f = 180 + 15/2 = 187.5 mm

web: A_w = 8·180 = 1440 mm², y_w = 180/2 = 90 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3900·187.5 + 1440·90)/5340 = 161.2 mm

Parallel-axis theorem

flange: 260·15³/12 + 3900·(187.5 − 161.2)² = 2.769 × 10⁶ mm⁴

web: 8·180³/12 + 1440·(90 − 161.2)² = 1.119 × 10⁷ mm⁴

I_x = 2.769 × 10⁶ + 1.119 × 10⁷ = 1.396 × 10⁷ mm⁴ = 13.96 × 10⁶ mm⁴

Answer

I_x = 13.96 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_j = 47.33 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 = 340 − 2·18 = 304 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (250·340³ − 240·304³)/12

I_x = 2.569 × 10⁸ mm⁴

Bending stress

y = 170 − 18 = 152 mm

σ = M·y/I = 8 × 10⁷·152/2.569 × 10⁸ = 47.33 MPa

Answer

σ_j = 47.33 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 46.1 MPa

Radius of Mohr’s circle

τ_max = R = √(((110 − 20)/2)² + 10²) = 46.1 MPa

Answer

τ_max = 46.1 MPa

10.Mohr’s circle & stress transformation

Answer: n = 4.55

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 = 55 MPa, R = 46.1 MPa

σ₁ = 101.1 MPa, σ₂ = 8.902 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 101.1 MPa

n = 460/101.1 = 4.55

Answer

n = 4.55

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

11.Beam deflection

Answer: δ = 18.75 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 = 10000 N, w = 2 N/mm, L = 3000 mm

EI = 5.88 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 10000·3000³/(3·5.88 × 10¹²) = 15.31 mm

δ_w = wL⁴/(8EI) = 2·3000⁴/(8·5.88 × 10¹²) = 3.444 mm

Add them

δ = 15.31 + 3.444 = 18.75 mm

Answer

δ = 18.75 mm

12.Beam deflection

Answer: M_A = 37.5 kN·m

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 = 3·12·5/8 = 22.5 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = wL²/2 − R_B·L = 150 − 112.5 = 37.5 kN·m

Answer

M_A = 37.5 kN·m

13.Column buckling

Answer: P_cr = 195.3 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·636173/1500² = 195340 N = 195.3 kN

Answer

P_cr = 195.3 kN

14.Column buckling

Answer: σ_cr = 87.73 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 = 0.5·3000 = 1500 mm

KL/r = 1500/10 = 150

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/150² = 87.73 MPa

Answer

σ_cr = 87.73 MPa

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