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 bar is made of a steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 50 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: brass, Ø50 mm, 300 mm, E = 100 GPaδ = 0.25 mm

P = kN

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 75 kNsegment 1: Ø40 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 0.5 m, brass (E = 100 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 80 mm, inner diameter 60 mm, length 2.5 m) is fixed at one end and twisted by a torque of 500 N·m at the other. Find the angle of twist φ in degrees.

T = 500 N·md_o = 80 mmd_i = 60 mmL = 2.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 60 mm is fixed at A. A torque of 1000 N·m acts at B, 2.5 m from A, and a torque of 1500 N·m in the same direction acts at the free end C, a further 2.5 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 60 mmT_B = 1000 N·mT_C = 1500 N·mAB = 2.5 mBC = 2.5 mG = 26 GPa

φ_C = °

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.

35 kN @ 4 m50 kN @ 2 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)

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.

25 kN @ 6 m50 kN @ 4 m10 kN/m, 0–5 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)

7.Centroid & second moment of areaMedium

A 150 mm × 150 mm plate has a circular hole of diameter 50 mm on its 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 mmd = 50 mmhole centre: y = 97.5 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 240 mm plate has a rectangular cut-out 80 mm wide and 20 mm high, centred on the vertical centre line with its centre 84 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 240 mmcut-out: 80 × 20 mmhole centre: y = 84 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −30 MPa, σᵧ = −50 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = −30 MPaσ_y = −50 MPaτ_xy = −50 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 80 MPa, σᵧ = 120 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.

σ_x = 80 MPaσ_y = 120 MPaτ_xy = −20 MPaθ = 60°

σ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 12 kN/mL = 2 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1 m from support B. Find the deflection δ_mid at midspan.

P = 22 kNb = 1 mL = 5 mE = 200 GPa (steel)I = 57 × 10⁶ mm⁴

δ_mid = mm

13.Column bucklingMedium

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

L = 2 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is pinned at both ends. It must carry 60 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 60 kNn = 2L = 4 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 39.67 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

A₁ = 314.2 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·314.2) = 4.775 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·1963) = 1.528 × 10⁻⁶ mm/N

Load

P = 0.25/6.303 × 10⁻⁶ = 39667 N = 39.67 kN

Answer

P = 39.67 kN

2.Axial stress & strain

Answer: σ_max = 663.1 MPa

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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 59.68 MPa

σ₂ = 663.1 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 663.1 MPa

Answer

σ_max = 663.1 MPa

3.Torsion of shafts

Answer: φ = 1.002 °

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(80⁴ − 60⁴)/32 = 2.749 × 10⁶ mm⁴

Angle of twist

T = 500 N·m = 500000 N·mm

L = 2.5 m = 2500 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 500000·2500/(26000·2.749 × 10⁶) = 0.01749 rad

φ = 0.01749·180/π = 1.002°

Answer

φ = 1.002°

4.Torsion of shafts

Answer: φ_C = 17.32 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 1500 N·m

T_AB = T_C + T_B = 1500 + 1000 = 2500 N·m

Twist of each segment

J = π·60⁴/32 = 1.272 × 10⁶ mm⁴

G = 26 GPa = 26000 MPa

φ_AB = 2.5 × 10⁶·2500/(26000·1.272 × 10⁶) = 0.1889 rad

φ_BC = 1.5 × 10⁶·2500/(26000·1.272 × 10⁶) = 0.1134 rad

Twist of end C

φ_C = 0.1889 + 0.1134 = 0.3023 rad → |φ_C| = 17.32°

Answer

φ_C = 17.32°

5.Shear force & bending moment

Answer: C (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 = 65 kN, R₂ = 20 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 = 130, M_C = 160 kN·m, M_D = 120, M_E = 80, M_F = 40, M_G = 0

Answer

Mark C: M_C = 160 kN·m

6.Shear force & bending moment

Answer: D (x = 3 m)

Support reactions

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

R₁ = 30 kN, R₂ = 95 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 = 25, M_C = 40, M_D = 45 kN·m, M_E = 40, M_F = −25, M_G = 0

Answer

Mark D: M_D = 45 kN·m

7.Centroid & second moment of area

Answer: I_x = 40.79 × 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 − 1963·97.5)/(22500 − 1963) = 72.85 mm

Full plate about ȳ

150·150³/12 + 22500·2.151² = 42.29 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·24.65² = 1.5 × 10⁶ mm⁴

Subtract

I_x = 40.79 × 10⁶ mm⁴

Answer

I_x = 40.79 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 136 × 10⁶ mm⁴

Centroid

ȳ = (28800·120 − 1600·84)/(28800 − 1600) = 122.1 mm

Full plate about ȳ

120·240³/12 + 28800·−2.118² = 138.4 × 10⁶ mm⁴

Hole about ȳ

80·20³/12 + 1600·−38.12² = 2.378 × 10⁶ mm⁴

Subtract

I_x = 136 × 10⁶ mm⁴

Answer

I_x = 136 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = −84.35 °

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

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −50/50.99 = −0.9806, sin 2θ_s = −10/50.99 = −0.1961

2θ_s = atan2(−10, −50) = −168.7°

Angle on the element

θ_s = −84.35°

check: θ_s − θ_p = −84.35 − (−39.35) = −45°

Answer

θ_s = −84.35°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 92.68 MPa

Transformation equation

σ_θ = 100 + −20·−0.5 + −20·0.866 = 92.68 MPa

Answer

σ_θ = 92.68 MPa

11.Beam deflection

Answer: δ_max = 1.894 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.

w = 12 N/mm, L = 2000 mm

EI = 1.32 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·12·2000⁴/(384·1.32 × 10¹²) = 1.894 mm

Answer

δ_max = 1.894 mm

12.Beam deflection

Answer: δ_mid = 2.855 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, b = 1000 mm, L = 5000 mm

EI = 1.14 × 10¹³ N·mm²

Midspan deflection

δ_mid = Pb(3L² − 4b²)/(48EI) = 22000·1000·(3·5000² − 4·1000²)/(48·1.14 × 10¹³) = 2.855 mm

Answer

δ_mid = 2.855 mm

13.Column buckling

Answer: P_cr = 248.1 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/4000² = 248050 N = 248.1 kN

Answer

P_cr = 248.1 kN

14.Column buckling

Answer: d = 86.74 mm

Required critical load

P_cr = n·P = 2·60000 = 120000 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·120000·4000²/(π³·70000))^(1/4) = 86.74 mm

Answer

d = 86.74 mm

Check: the slenderness KL/r is 184.5, so the column is slender and Euler applies.

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