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 600 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 300 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 300 mm, E = 110 GPaδ = 0.3 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 = 100 kNsegment 1: Ø35 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, steel (E = 200 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 250 mm rotate 0.57° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 50 N·md = 20 mmL = 250 mmφ = 0.57°

G = GPa

4.Torsion of shaftsHard

A hollow shaft must transmit 30 kW at 300 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

P = 30 kWN = 300 rpmd_i/d_o = 0.75τ_allow = 75 MPa

d_o = 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.

40 kN @ 3 m10 kN/m, 0–6 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.

55 kN @ 0 m45 kN @ 6 m20 kN @ 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)

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 60 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 = 60 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 5 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 200 mmt_f = 30 mmt_w = 8 mmh_w = 120 mmM = 5 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 20 MPa, σᵧ = −50 MPa and τₓᵧ = 30 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 = −50 MPaτ_xy = 30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −80 MPa, σᵧ = −60 MPa and τₓᵧ = −50 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 = −60 MPaτ_xy = −50 MPaθ = 60°

σ_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 = 15 kN/mL = 4 mE = 70 GPa (aluminium)I = 23 × 10⁶ mm⁴

θ_A = rad

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the deflection δ_mid at midspan.

P = 28 kNb = 1.5 mL = 5 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴

δ_mid = 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 = 6 mE = 70 GPa (aluminium)b × h = 25 × 50 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 = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 27.43 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) = 600/(200000·314.2) = 9.549 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(110000·1963) = 1.389 × 10⁻⁶ mm/N

Load

P = 0.3/1.094 × 10⁻⁵ = 27427 N = 27.43 kN

Answer

P = 27.43 kN

2.Axial stress & strain

Answer: σ_max = 884.2 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·35²/4 = 962.1 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 103.9 MPa

σ₂ = 884.2 MPa

Largest stress

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

σ_max = 884.2 MPa

Answer

σ_max = 884.2 MPa

3.Torsion of shafts

Answer: G = 79.99 GPa

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 moment and angle

J = π·20⁴/32 = 15708 mm⁴

φ = 0.57·π/180 = 0.009948 rad

Shear modulus

T = 50 N·m = 50000 N·mm

G = TL/(Jφ) = 50000·250/(15708·0.009948) = 79990 MPa

G = 79.99 GPa

Answer

G = 79.99 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d_o = 45.61 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 30000/31.42 = 954.9 N·m

Outer diameter from the allowable stress

T = 954.9 N·m = 954930 N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·954930/(π·75·0.6836)) = 45.61 mm

d_i = 0.75·45.61 = 34.2 mm

Answer

d_o = 45.61 mm

A solid shaft would need d = 40.18 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: D (x = 3 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 = 50 kN, R₂ = 50 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 = 45, M_C = 80, M_D = 105 kN·m, M_E = 80, M_F = 45, M_G = 0

Answer

Mark D: M_D = 105 kN·m

6.Shear force & bending moment

Answer: B (x = 1 m)

Support reactions

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

R₁ = 62.5 kN, R₂ = 57.5 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 = −55 kN·m, M_C = −47.5, M_D = −40, M_E = −32.5, M_F = −45, M_G = 0

Answer

Mark B: M_B = −55 kN·m

7.Centroid & second moment of area

Answer: I_x = 56.05 × 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 − 706.9·60)/(30000 − 706.9) = 75.36 mm

Full plate about ȳ

200·150³/12 + 30000·−0.362² = 56.25 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−15.36² = 0.2066 × 10⁶ mm⁴

Subtract

I_x = 56.05 × 10⁶ mm⁴

Answer

I_x = 56.05 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 99.61 MPa

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 two rectangles

Measure every y from the bottom edge.

flange: A_f = 200·30 = 6000 mm², y_f = 120 + 30/2 = 135 mm

web: A_w = 8·120 = 960 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·135 + 960·60)/6960 = 124.7 mm

Parallel-axis theorem

flange: 200·30³/12 + 6000·(135 − 124.7)² = 1.092 × 10⁶ mm⁴

web: 8·120³/12 + 960·(60 − 124.7)² = 5.165 × 10⁶ mm⁴

I_x = 1.092 × 10⁶ + 5.165 × 10⁶ = 6.257 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 124.7 mm

σ = M·y/I = 5 × 10⁶·124.7/6.257 × 10⁶ = 99.61 MPa

Answer

σ_bottom = 99.61 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 20.3 °

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(2·30, 20 − −50) = 40.6°

Angle on the element

θ_p = 20.3°

Answer

θ_p = 20.3°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = −108.3 MPa

Transformation equation

σ_θ = −70 + −10·−0.5 + −50·0.866 = −108.3 MPa

Answer

σ_θ = −108.3 MPa

11.Beam deflection

Answer: θ_A = 0.02484 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 = 15 N/mm, L = 4000 mm

EI = 1.61 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 15·4000³/(24·1.61 × 10¹²) = 0.02484 rad

Answer

θ_A = 0.02484 rad

That is 1.423°.

12.Beam deflection

Answer: δ_mid = 5.156 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 = 28000 N, b = 1500 mm, L = 5000 mm

EI = 1.12 × 10¹³ N·mm²

Midspan deflection

δ_mid = Pb(3L² − 4b²)/(48EI) = 28000·1500·(3·5000² − 4·1500²)/(48·1.12 × 10¹³) = 5.156 mm

Answer

δ_mid = 5.156 mm

13.Column buckling

Answer: P_cr = 2.55 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 = 50·25³/12 = 65104 mm⁴

Effective length

KL = 0.7·6000 = 4200 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·65104/4200² = 2550 N = 2.55 kN

Answer

P_cr = 2.55 kN

14.Column buckling

Answer: σ_cr = 49.35 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/7.5 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa

Answer

σ_cr = 49.35 MPa

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