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 steel bar under the axial load.

P = 15 kNd = 16 mmL = 0.5 mE = 200 GPa

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø120/Ø50 mm, E = 120 GPa, σ_allow = 70 MPa

P_max = kN

3.Torsion of shaftsMedium

A solid steel 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 = 100 N·md = 20 mmL = 1.2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 300 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 10 kWN = 300 rpmτ_allow = 50 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 @ 2 m6 kN/m, 2–4 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

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.

10 kN @ 12 m55 kN @ 8 m8 kN/m, 0–10 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

7.Centroid & second moment of areaMedium

A 200 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 105 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 150 mmd = 30 mmhole centre: y = 105 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 160 × 20 mm, a web 210 mm high and 12 mm thick, and a bottom flange 300 × 16 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 160 × 20 mmweb: 12 × 210 mmbottom flange: 300 × 16 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −40 MPa, σᵧ = 100 MPa and τₓᵧ = −60 MPa (tension positive). The element is rotated 20° 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 = −40 MPaσ_y = 100 MPaτ_xy = −60 MPaθ = 20°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 15 mm holds gas at a gauge pressure of 1 MPa. It is made from plate welded along a helical seam at 35° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 1500 mmt = 15 mmp = 1 MPaα = 35°

|τ_n| = MPa

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 = 28 kNw = 4 kN/mL = 3 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ = 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 = 10 kN/mL = 5 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 1.5 mE = 200 GPa (steel)d = 40 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 = 20 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.1865 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·16²/4 = 201.1 mm²

σ = P/A = 15·1000/201.1 = 74.6 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 74.6/200000 = 0.000373 = 373 µε

Elongation

δ = ε·L = 0.000373·500 mm = 0.1865 mm

Answer

δ = 0.1865 mm

2.Axial stress & strain

Answer: P_max = 711.5 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·25²/4 = 490.9 mm²

A_t = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 120·10³·9346 = 1.22 × 10⁹ N

Governing strain

ε = min(0.001, 0.0005833) = 0.0005833 (copper)

Largest load

P_max = 0.0005833·1.22 × 10⁹ = 711505 N = 711.5 kN

Answer

P_max = 711.5 kN

3.Torsion of shafts

Answer: φ = 5.471 °

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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1200/(80000·15708) = 0.09549 rad

φ = 0.09549·180/π = 5.471°

Answer

φ = 5.471°

4.Torsion of shafts

Answer: d = 31.89 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 10000/31.42 = 318.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·318310/(π·50)) = 31.89 mm

Answer

d = 31.89 mm

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

5.Shear force & bending moment

Answer: C (x = 2 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 = 42.67 kN, R₂ = 24.33 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 = 42.67, M_C = 85.33 kN·m, M_D = 70, M_E = 48.67, M_F = 24.33, M_G = 0

Answer

Mark C: M_C = 85.33 kN·m

6.Shear force & bending moment

Answer: D (x = 6 m)

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 49 kN, R₂ = 96 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 = 82, M_C = 132, M_D = 150 kN·m, M_E = 136, M_F = −20, M_G = 0

Answer

Mark D: M_D = 150 kN·m

7.Centroid & second moment of area

Answer: I_x = 55.56 × 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 = (30000·75 − 706.9·105)/(30000 − 706.9) = 74.28 mm

Full plate about ȳ

200·150³/12 + 30000·0.7239² = 56.27 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·30.72² = 0.707 × 10⁶ mm⁴

Subtract

I_x = 55.56 × 10⁶ mm⁴

Answer

I_x = 55.56 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 110.2 × 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 three rectangles

Measure every y from the bottom edge.

top flange: A = 3200 mm², y = 236 mm

web: A = 2520 mm², y = 121 mm

bottom flange: A = 4800 mm², y = 8 mm

Centroid

ȳ = (3200·236 + 2520·121 + 4800·8)/10520 = 104.4 mm

Parallel-axis theorem

top flange: 160·20³/12 + 3200·131.6² = 5.551 × 10⁷ mm⁴

web: 12·210³/12 + 2520·16.58² = 9.954 × 10⁶ mm⁴

bottom flange: 300·16³/12 + 4800·96.42² = 4.473 × 10⁷ mm⁴

I_x = 5.551 × 10⁷ + 9.954 × 10⁶ + 4.473 × 10⁷

I_x = 1.102 × 10⁸ mm⁴ = 110.2 × 10⁶ mm⁴

Answer

I_x = 110.2 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 122.2 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′ = 30 − (−70)·0.766 − (−60)·0.6428 = 122.2 MPa

Check

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

σ_x′ = −62.19 MPa, σ_x′ + σ_y′ = 60 = σ_x + σ_y

Answer

σ_y′ = 122.2 MPa

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 11.75 MPa

Wall stresses

r = 1500/2 = 750 mm

σ_h = pr/t = 1·750/15 = 50 MPa

σ_a = pr/(2t) = 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| = (50 − 25)·sin 35°·cos 35° = 11.75 MPa

Answer

|τ_n| = 11.75 MPa

11.Beam deflection

Answer: δ = 9.141 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 = 28000 N, w = 4 N/mm, L = 3000 mm

EI = 3.2 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(3EI) = 28000·3000³/(3·3.2 × 10¹³) = 7.875 mm

δ_w = wL⁴/(8EI) = 4·3000⁴/(8·3.2 × 10¹³) = 1.266 mm

Add them

δ = 7.875 + 1.266 = 9.141 mm

Answer

δ = 9.141 mm

12.Beam deflection

Answer: I_req = 29.3 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 5000/360 = 13.89 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·10·5000⁴/(384·200000·13.89) = 2.93 × 10⁷ mm⁴

I_req = 29.3 × 10⁶ mm⁴

Answer

I_req = 29.3 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 225 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·125664/1050² = 224989 N = 225 kN

Answer

P_cr = 225 kN

14.Column buckling

Answer: σ_cr = 21.93 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/5 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/300² = 21.93 MPa

Answer

σ_cr = 21.93 MPa

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