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 axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).

P = 120 kNd = 20 mmL = 2.5 mE = 70 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø100/Ø60 mm, E = 120 GPa, σ_allow = 50 MPa

P_max = kN

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 2000 N·m. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

T = 2000 N·md_i/d_o = 0.8τ_allow = 75 MPa

d_o = mm

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 3000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 20 kWN = 3000 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 8 kN/m from 0 to 5 mx = 1.5 m

V_x = kN

6.Shear force & bending momentHard

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 = 5 mP = 15 kN at 4 mw = 12 kN/m from 0 to 5 mx = 4 m

M_x = kN·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 = 25 mmt_w = 8 mmh_w = 100 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 100 mm × 240 mm plate has a rectangular cut-out 30 mm wide and 40 mm high, centred on the vertical centre line with its centre 168 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 240 mmcut-out: 30 × 40 mmhole centre: y = 168 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −80 MPa, σᵧ = 110 MPa and τₓᵧ = 40 MPa (tension positive). The element is rotated 25° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = −80 MPaσ_y = 110 MPaτ_xy = 40 MPaθ = 25°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −20 MPa, σᵧ = −50 MPa and τₓᵧ = −60 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 30° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −20 MPaσ_y = −50 MPaτ_xy = −60 MPaθ = -30°

τ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 2.5 m from A. Find the deflection δ_B at the free end B.

P = 26 kNa = 2.5 mL = 5 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ_B = 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 steel is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 200 GPa (steel)d = 20 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 = 60 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 5457 µε

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·20²/4 = 314.2 mm²

σ = P/A = 120·1000/314.2 = 382 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 382/70000 = 0.005457 = 5457 µε

Answer

ε = 5457 µε

2.Axial stress & strain

Answer: P_max = 356 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

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·5027 = 8.545 × 10⁸ N

Governing strain

ε = min(0.0006, 0.0004167) = 0.0004167 (copper)

Largest load

P_max = 0.0004167·8.545 × 10⁸ = 356047 N = 356 kN

Answer

P_max = 356 kN

3.Torsion of shafts

Answer: d_o = 61.27 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.

Outer diameter from the allowable stress

T = 2000 N·m = 2 × 10⁶ N·mm

1 − k⁴ = 1 − 0.8⁴ = 0.5904

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·2 × 10⁶/(π·75·0.5904)) = 61.27 mm

d_i = 0.8·61.27 = 49.02 mm

Answer

d_o = 61.27 mm

A solid shaft would need d = 51.4 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 17.55 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 20000/314.2 = 63.66 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·63662/(π·60)) = 17.55 mm

Answer

d = 17.55 mm

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

5.Shear force & bending moment

Answer: V_x = 8 kN

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 = 8·5 = 40 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 8·5·2.5

R_B = 100/5 = 20 kN

ΣF_y = 0: R_A = 40 − 20 = 20 kN

Cut at x = 1.5 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.

V = R_A − (all downward loads left of the cut)

V_x = 20 − 8·1.5 = 8 kN

Answer

V_x = 8 kN

6.Shear force & bending moment

Answer: M_x = 36 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·5 = 60 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 15·4 + 12·5·2.5

R_B = 210/5 = 42 kN

ΣF_y = 0: R_A = 75 − 42 = 33 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 = 33·4 − 12·4·2 = 36 kN·m

Answer

M_x = 36 kN·m

7.Centroid & second moment of area

Answer: I_x = 3.788 × 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 = 260·25 = 6500 mm², y_f = 100 + 25/2 = 112.5 mm

web: A_w = 8·100 = 800 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6500·112.5 + 800·50)/7300 = 105.7 mm

Parallel-axis theorem

flange: 260·25³/12 + 6500·(112.5 − 105.7)² = 643477 mm⁴

web: 8·100³/12 + 800·(50 − 105.7)² = 3.144 × 10⁶ mm⁴

I_x = 643477 + 3.144 × 10⁶ = 3.788 × 10⁶ mm⁴ = 3.788 × 10⁶ mm⁴

Answer

I_x = 3.788 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 112.1 × 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 = (24000·120 − 1200·168)/(24000 − 1200) = 117.5 mm

Full plate about ȳ

100·240³/12 + 24000·2.526² = 115.4 × 10⁶ mm⁴

Hole about ȳ

30·40³/12 + 1200·50.53² = 3.223 × 10⁶ mm⁴

Subtract

I_x = 112.1 × 10⁶ mm⁴

Answer

I_x = 112.1 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 45.42 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.

Transformation equation for the y′ face

σ_y′ = 15 − (−95)·0.6428 − 40·0.766 = 45.42 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = −15.42 MPa, σ_x′ + σ_y′ = 30 = σ_x + σ_y

Answer

σ_y′ = 45.42 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = −17.01 MPa

Transformation equation

τ_θ = −15·−0.866 + −60·0.5 = −17.01 MPa

Answer

τ_θ = −17.01 MPa

11.Beam deflection

Answer: δ_B = 10.58 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 = 26000 N, a = 2500 mm, L = 5000 mm

EI = 3.2 × 10¹³ N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 4.232 mm

θ_P = Pa²/(2EI) = 0.002539 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 10.58 mm

Answer

δ_B = 10.58 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 = 62.01 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·20⁴/64 = 7854 mm⁴

Effective length

KL = 0.5·1000 = 500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·7854/500² = 62013 N = 62.01 kN

Answer

P_cr = 62.01 kN

14.Column buckling

Answer: σ_cr = 197.4 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/15 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/100² = 197.4 MPa

Answer

σ_cr = 197.4 MPa

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