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 = 45 kNd = 10 mmL = 1.5 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 brass (outer diameter 120 mm, inner diameter 60 mm, E = 100 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø120/Ø60 mm, E = 100 GPa, σ_allow = 60 MPa

P_max = kN

3.Torsion of shaftsMedium

A stepped solid steel shaft is fixed at A. Segment AB has diameter 40 mm and length 1 m; segment BC has diameter 30 mm and length 0.5 m. A torque of 400 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 400 N·md₁ = 40 mm, L₁ = 1 md₂ = 30 mm, L₂ = 0.5 mG = 80 GPa

φ_C = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 1800 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 50 kWN = 1800 rpmτ_allow = 75 MPa

d = mm

5.Shear force & bending momentMedium

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.

55 kN @ 4 m12 kN/m, 1–5 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)

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 15 kN at 3.5 mw = 5 kN/m from 0 to 5 mx = 3 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 280 mm wide with two upright legs, 170 mm high overall. Base and legs are all 8 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 280 mmh = 170 mmt = 8 mm

ȳ = mm

8.Centroid & second moment of areaHard

A 150 mm × 150 mm plate has a rectangular cut-out 80 mm wide and 30 mm high, centred on the vertical centre line with its centre 97.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 150 mmcut-out: 80 × 30 mmhole centre: y = 97.5 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −20 MPa, σᵧ = 0 MPa and τₓᵧ = 40 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = −20 MPaσ_y = 0 MPaτ_xy = 40 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 6 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 30° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 500 mmt = 6 mmp = 1.5 MPaα = 30°

|τ_n| = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 3.5 m from support A. Find the deflection δ_P under the load.

P = 3 kNa = 3.5 mL = 8 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 22 kNw = 6 kN/mL = 5 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 70 GPa (aluminium)b × h = 20 × 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 70 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 70 kNn = 3L = 2 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 8185 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·10²/4 = 78.54 mm²

σ = P/A = 45·1000/78.54 = 573 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 573/70000 = 0.008185 = 8185 µε

Answer

ε = 8185 µε

2.Axial stress & strain

Answer: P_max = 659.7 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(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 100·10³·8482 = 1.1 × 10⁹ N

Governing strain

ε = min(0.0006, 0.0006) = 0.0006 (brass)

Largest load

P_max = 0.0006·1.1 × 10⁹ = 659734 N = 659.7 kN

Answer

P_max = 659.7 kN

3.Torsion of shafts

Answer: φ_C = 2.941 °

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 moments

J₁ = π·40⁴/32 = 251327 mm⁴

J₂ = π·30⁴/32 = 79522 mm⁴

Twist of each segment

T = 400 N·m = 400000 N·mm

G = 80 GPa = 80000 MPa

φ_AB = 400000·1000/(80000·251327) = 0.01989 rad

φ_BC = 400000·500/(80000·79522) = 0.03144 rad

Total twist

φ_C = 0.01989 + 0.03144 = 0.05133 rad = 2.941°

Answer

φ_C = 2.941°

4.Torsion of shafts

Answer: d = 26.21 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 50000/188.5 = 265.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·265258/(π·75)) = 26.21 mm

Answer

d = 26.21 mm

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

5.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 = 42.33 kN, R₂ = 60.67 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 = 42.33, M_C = 78.67, M_D = 103, M_E = 115.3 kN·m, M_F = 60.67, M_G = 0

Answer

Mark E: M_E = 115.3 kN·m

6.Shear force & bending moment

Answer: M_x = 28.5 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·5 = 25 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 15·3.5 + 5·5·2.5

R_B = 115/5 = 23 kN

ΣF_y = 0: R_A = 40 − 23 = 17 kN

Cut at x = 3 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 = 17·3 − 5·3·1.5 = 28.5 kN·m

Answer

M_x = 28.5 kN·m

7.Centroid & second moment of area

Answer: ȳ = 49.6 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 = 280·8 = 2240 mm², y₁ = 4 mm

two legs: A₂ = 2·8·162 = 2592 mm², y₂ = 8 + 162/2 = 89 mm

Centroid

ȳ = (2240·4 + 2592·89)/4832 = 49.6 mm

Answer

ȳ = 49.6 mm

8.Centroid & second moment of area

Answer: I_x = 40.65 × 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 = (22500·75 − 2400·97.5)/(22500 − 2400) = 72.31 mm

Full plate about ȳ

150·150³/12 + 22500·2.687² = 42.35 × 10⁶ mm⁴

Hole about ȳ

80·30³/12 + 2400·25.19² = 1.702 × 10⁶ mm⁴

Subtract

I_x = 40.65 × 10⁶ mm⁴

Answer

I_x = 40.65 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = 52.02 °

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.

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·40, −20 − 0) = 104°

Angle on the element

θ_p = 52.02°

Answer

θ_p = 52.02°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 13.53 MPa

Wall stresses

r = 500/2 = 250 mm

σ_h = pr/t = 1.5·250/6 = 62.5 MPa

σ_a = pr/(2t) = 31.25 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

|τ_n| = (62.5 − 31.25)·sin 30°·cos 30° = 13.53 MPa

Answer

|τ_n| = 13.53 MPa

11.Beam deflection

Answer: δ_P = 2.769 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 = 3000 N, a = 3500 mm, b = 4500 mm, L = 8000 mm

EI = 1.12 × 10¹³ N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 3000·3500²·4500²/(3·1.12 × 10¹³·8000) = 2.769 mm

Answer

δ_P = 2.769 mm

12.Beam deflection

Answer: δ_max = 26.6 mm

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 = 22000 N, w = 6 N/mm, L = 5000 mm, EI = 3.99 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 22000·5000³/(48·3.99 × 10¹²) = 14.36 mm

δ_w = 5wL⁴/(384EI) = 5·6·5000⁴/(384·3.99 × 10¹²) = 12.24 mm

Add them

δ_max = 14.36 + 12.24 = 26.6 mm

Answer

δ_max = 26.6 mm

13.Column buckling

Answer: P_cr = 37.6 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 = 40·20³/12 = 26667 mm⁴

Effective length

KL = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·26667/700² = 37598 N = 37.6 kN

Answer

P_cr = 37.6 kN

14.Column buckling

Answer: d = 38.37 mm

Required critical load

P_cr = n·P = 3·70000 = 210000 N

KL = 0.5·2000 = 1000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·210000·1000²/(π³·200000))^(1/4) = 38.37 mm

Answer

d = 38.37 mm

Check: the slenderness KL/r is 104.3, so the column is slender and Euler applies.

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