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

P = 95 kNd = 10 mmL = 2.5 mE = 100 GPa

ε = µε

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 = 40 kNsegment 1: Ø40 mm, 1 m, brass (E = 100 GPa)segment 2: Ø16 mm, 2 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 50 mm, inner diameter 25 mm, length 2.5 m) is fixed at one end and twisted by a torque of 100 N·m at the other. Find the angle of twist φ in degrees.

T = 100 N·md_o = 50 mmd_i = 25 mmL = 2.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft of diameter 60 mm is fixed at A. A torque of 2000 N·m acts at B, 1.2 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 0.5 m along. Find the maximum shear stress τ_max in the shaft.

d = 60 mmT_B = 2000 N·mT_C = 250 N·mAB = 1.2 mBC = 0.5 m

τ_max = MPa

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.

45 kN @ 6 m10 kN/m, 0–12 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

Find the bending moment M_x at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 60 kN at 9 mw = 10 kN/m from 0 to 10 mx = 4.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A symmetric I-section has flanges 240 mm wide and 14 mm thick, an overall depth of 380 mm and a web 6 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 240 mmH = 380 mmt_f = 14 mmt_w = 6 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 200 mm plate has a rectangular cut-out 50 mm wide and 40 mm high, centred on the vertical centre line with its centre 120 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 200 mmcut-out: 50 × 40 mmhole centre: y = 120 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −60 MPa, σᵧ = −10 MPa and τₓᵧ = 50 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 = −60 MPaσ_y = −10 MPaτ_xy = 50 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 120 MPa and τₓᵧ = 60 MPa (σ_z = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 120 MPaσ_y = 120 MPaτ_xy = 60 MPaσ_Y = 460 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 5 m from support A. Find the deflection δ_P under the load.

P = 4 kNa = 5 mL = 6 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 75 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/300. Find the smallest depth h_req of the section.

w = 1 kN/mL = 3 mE = 11 GPa (timber)b = 75 mm

h_req = mm

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 = 5 mE = 200 GPa (steel)b × h = 30 × 60 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 110 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 110 kNn = 2.5L = 5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 12096 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 = 95·1000/78.54 = 1210 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 1210/100000 = 0.0121 = 12096 µε

Answer

ε = 12096 µε

2.Axial stress & strain

Answer: σ_max = 198.9 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·16²/4 = 201.1 mm²

σ₁ = 31.83 MPa

σ₂ = 198.9 MPa

Largest stress

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

σ_max = 198.9 MPa

Answer

σ_max = 198.9 MPa

3.Torsion of shafts

Answer: φ = 0.9577 °

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(50⁴ − 25⁴)/32 = 575243 mm⁴

Angle of twist

T = 100 N·m = 100000 N·mm

L = 2.5 m = 2500 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 100000·2500/(26000·575243) = 0.01672 rad

φ = 0.01672·180/π = 0.9577°

Answer

φ = 0.9577°

4.Torsion of shafts

Answer: τ_max = 53.05 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 250 N·m

T_AB = T_C + T_B = 250 + 2000 = 2250 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 2250 N·m

τ_max = 16·2.25 × 10⁶/(π·60³) = 53.05 MPa

Answer

τ_max = 53.05 MPa

5.Shear force & bending moment

Answer: D (x = 6 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 = 82.5 kN, R₂ = 82.5 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 = 145, M_C = 250, M_D = 315 kN·m, M_E = 250, M_F = 145, M_G = 0

Answer

Mark D: M_D = 315 kN·m

6.Shear force & bending moment

Answer: M_x = 150.8 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 = 10·10 = 100 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 60·9 + 10·10·5

R_B = 1040/10 = 104 kN

ΣF_y = 0: R_A = 160 − 104 = 56 kN

Cut at x = 4.5 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 = 56·4.5 − 10·4.5·2.25 = 150.8 kN·m

Answer

M_x = 150.8 kN·m

7.Centroid & second moment of area

Answer: I_x = 247 × 10⁶ mm⁴

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.

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 380 − 2·14 = 352 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (240·380³ − 234·352³)/12

I_x = 2.47 × 10⁸ mm⁴ = 247 × 10⁶ mm⁴

Answer

I_x = 247 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 98.88 × 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 = (30000·100 − 2000·120)/(30000 − 2000) = 98.57 mm

Full plate about ȳ

150·200³/12 + 30000·1.429² = 100.1 × 10⁶ mm⁴

Hole about ȳ

50·40³/12 + 2000·21.43² = 1.185 × 10⁶ mm⁴

Subtract

I_x = 98.88 × 10⁶ mm⁴

Answer

I_x = 98.88 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = 58.28 °

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(2·50, −60 − −10) = 116.6°

Angle on the element

θ_p = 58.28°

Answer

θ_p = 58.28°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 2.898

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 120 MPa, R = 60 MPa

σ₁ = 180 MPa, σ₂ = 60 MPa, σ₃ = 0

Von Mises

σ_vM = √(180² − 180·60 + 60²) = 158.7 MPa

n = 460/158.7 = 2.898

Answer

n = 2.898

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ_P = 2.205 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 = 4000 N, a = 5000 mm, b = 1000 mm, L = 6000 mm

EI = 2.52 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 4000·5000²·1000²/(3·2.52 × 10¹²·6000) = 2.205 mm

Answer

δ_P = 2.205 mm

12.Beam deflection

Answer: h_req = 115.3 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.

w = 1 N/mm, L = 3000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/300 = 3000/300 = 10 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·1·3000⁴/(384·11000·10) = 9.588 × 10⁶ mm⁴

Required depth

bh³/12 = I_req → h = (12·9.588 × 10⁶/75)^1/3 = 115.3 mm

Answer

h_req = 115.3 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 42.64 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·135000/2500² = 42637 N = 42.64 kN

Answer

P_cr = 42.64 kN

14.Column buckling

Answer: d = 64.9 mm

Required critical load

P_cr = n·P = 2.5·110000 = 275000 N

KL = 0.5·5000 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·275000·2500²/(π³·200000))^(1/4) = 64.9 mm

Answer

d = 64.9 mm

Check: the slenderness KL/r is 154.1, so the column is slender and Euler applies.

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