Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 steel bar, in microstrain (µε = 10⁻⁶).

P = 95 kNd = 20 mmL = 2.5 mE = 200 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø120/Ø60 mm, E = 120 GPa, σ_allow = 100 MPa

P_max = kN

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 = 1000 N·md = 40 mmL = 0.8 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 50 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 50 kWN = 300 rpmd_o = 100 mmd_i = 75 mm

τ_max = MPa

5.Shear force & bending momentMedium

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. It carries a point load P₁ between the supports and a point load P₂ at C. Find the bending moment M_x under the load P₁. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP₁ = 75 kN at 2 mP₂ = 65 kN at 6 msupports at x = 0 and x = 5 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 5.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 55 kN at 5 mw = 2 kN/m from 0 to 6 mx = 5.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 170 mm and inside diameter d = 154 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 170 mmd = 154 mm

r = mm

8.Centroid & second moment of areaHard

A 100 mm × 240 mm plate has a rectangular cut-out 60 mm wide and 30 mm high, centred on the vertical centre line with its centre 72 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: 60 × 30 mmhole centre: y = 72 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 110 MPa, σᵧ = 90 MPa and τₓᵧ = 10 MPa (tension positive). The element is rotated 15° 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 = 110 MPaσ_y = 90 MPaτ_xy = 10 MPaθ = 15°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 60 MPa, σᵧ = −70 MPa and τₓᵧ = 10 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 30° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 60 MPaσ_y = −70 MPaτ_xy = 10 MPaθ = 30°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 8 kN/mL = 3 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

θ_A = rad

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 = 5 kN/mL = 4 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

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 = 2 mE = 200 GPa (steel)d = 30 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 = 5 mE = 200 GPa (steel)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 1512 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 = 95·1000/314.2 = 302.4 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 302.4/200000 = 0.001512 = 1512 µε

Answer

ε = 1512 µε

2.Axial stress & strain

Answer: P_max = 966 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·30²/4 = 706.9 mm²

A_t = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·8482 = 1.159 × 10⁹ N

Governing strain

ε = min(0.00125, 0.0008333) = 0.0008333 (copper)

Largest load

P_max = 0.0008333·1.159 × 10⁹ = 966040 N = 966 kN

Answer

P_max = 966 kN

3.Torsion of shafts

Answer: φ = 7.015 °

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) = 1000·1000·800/(26000·251327) = 0.1224 rad

φ = 0.1224·180/π = 7.015°

Answer

φ = 7.015°

4.Torsion of shafts

Answer: τ_max = 11.86 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 50000/31.42 = 1592 N·m

Polar second moment of area

J = π(100⁴ − 75⁴)/32 = 6.711 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 1.592 × 10⁶·50/6.711 × 10⁶ = 11.86 MPa

Answer

τ_max = 11.86 MPa

5.Shear force & bending moment

Answer: M_x = 64 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. The overhang load also acts about A.

ΣM_A = 0: R_B·5 = 75·2 + 65·6

R_B = 540/5 = 108 kN

ΣF_y = 0: R_A = 140 − 108 = 32 kN

Cut under P₁ and keep the left part

M_x = R_A·2 = 32·2 = 64 kN·m

Answer

M_x = 64 kN·m

6.Shear force & bending moment

Answer: M_x = 25.67 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 2·6 = 12 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 55·5 + 2·6·3

R_B = 311/6 = 51.83 kN

ΣF_y = 0: R_A = 67 − 51.83 = 15.17 kN

Cut at x = 5.5 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 = 15.17·5.5 − 55·(5.5 − 5) − 2·5.5·2.75 = 25.67 kN·m

Answer

M_x = 25.67 kN·m

7.Centroid & second moment of area

Answer: r = 57.35 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 4072 mm²

I = π(D⁴ − d⁴)/64 = 1.339 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(1.339 × 10⁷/4072) = 57.35 mm

Answer

r = 57.35 mm

8.Centroid & second moment of area

Answer: I_x = 110.6 × 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 − 1800·72)/(24000 − 1800) = 123.9 mm

Full plate about ȳ

100·240³/12 + 24000·−3.892² = 115.6 × 10⁶ mm⁴

Hole about ȳ

60·30³/12 + 1800·−51.89² = 4.982 × 10⁶ mm⁴

Subtract

I_x = 110.6 × 10⁶ mm⁴

Answer

I_x = 110.6 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 86.34 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′ = 100 − 10·0.866 − 10·0.5 = 86.34 MPa

Check

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

σ_x′ = 113.7 MPa, σ_x′ + σ_y′ = 200 = σ_x + σ_y

Answer

σ_y′ = 86.34 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = −51.29 MPa

Transformation equation

τ_θ = −65·0.866 + 10·0.5 = −51.29 MPa

Answer

τ_θ = −51.29 MPa

11.Beam deflection

Answer: θ_A = 0.005114 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 8 N/mm, L = 3000 mm

EI = 1.76 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 8·3000³/(24·1.76 × 10¹²) = 0.005114 rad

Answer

θ_A = 0.005114 rad

That is 0.293°.

12.Beam deflection

Answer: I_req = 7.5 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·5·4000⁴/(384·200000·11.11) = 7.5 × 10⁶ mm⁴

I_req = 7.5 × 10⁶ mm⁴

Answer

I_req = 7.5 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 78.48 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·2000 = 1000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·39761/1000² = 78485 N = 78.48 kN

Answer

P_cr = 78.48 kN

14.Column buckling

Answer: σ_cr = 126.3 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 0.5·5000 = 2500 mm

KL/r = 2500/20 = 125

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/125² = 126.3 MPa

Answer

σ_cr = 126.3 MPa

ExamProblems · examproblems.com/courses/mechanics-of-materials

Mechanics of Materials: cram pack

1Formulas and methods

Axial stress & strain

Torsion of shafts

Method

Shear force & bending moment

Method

Centroid & second moment of area

Mohr’s circle & stress transformation

Method

Beam deflection

Column buckling

2Practice problems

3Solutions

Stress

Strain (Hooke’s law)

Answer

Areas

Combined axial stiffness

Governing strain

Largest load

Answer

Polar second moment of area

Angle of twist

Answer

Torque

Polar second moment of area

Maximum shear stress

Answer

Support reactions

Cut under P1 and keep the left part

Answer

Support reactions

Cut at x = 5.5 m and keep the left part

Answer

Net area and second moment of area

Radius of gyration

Answer

Centroid

Full plate about ȳ

Hole about ȳ

Subtract

Answer

Transformation equation for the y′ face

Check

Answer

Transformation equation

Answer

Consistent units

End slope

Answer

Deflection limit

Required second moment of area

Answer

Smallest second moment of area

Effective length

Critical load

Answer

Smallest second moment of area

Radius of gyration and slenderness

Critical stress

Answer

Cut under P₁ and keep the left part