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

P = 60 kNd = 16 mmL = 1.5 mE = 200 GPa

ε = µε

2.Axial stress & strainHard

A steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaP = 80 kN

σ₂ = MPa

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 = 400 N·md = 30 mmL = 0.8 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 1000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 1000 rpmd_o = 40 mmd_i = 30 mm

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries two point loads. Find the maximum bending moment M_max in the beam.

L = 4 mP₁ = 80 kN at 1.5 mP₂ = 45 kN at 2.5 m

M_max = kN·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 = 5 mP = 60 kN at 2.5 mw = 10 kN/m from 0 to 5 mx = 4.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 120 mmt_f = 15 mmt_w = 20 mmh_w = 220 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 size of the bending stress σ_top at the top edge.

b_f = 180 mmt_f = 30 mmt_w = 12 mmh_w = 200 mmM = 5 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 80 MPa, σᵧ = 110 MPa and τₓᵧ = 50 MPa (tension positive). The element is rotated 50° 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 = 80 MPaσ_y = 110 MPaτ_xy = 50 MPaθ = 50°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −10 MPa, σᵧ = 100 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −10 MPaσ_y = 100 MPaτ_xy = 30 MPaθ = 15°

τ_θ = 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 = 29 kNw = 8 kN/mL = 2 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 5 mw = 15 kN/m

R_B = kN

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 = 1 mE = 200 GPa (steel)d = 20 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 40 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 40 kNn = 2L = 5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 1492 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·16²/4 = 201.1 mm²

σ = P/A = 60·1000/201.1 = 298.4 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 298.4/200000 = 0.001492 = 1492 µε

Answer

ε = 1492 µε

2.Axial stress & strain

Answer: σ₂ = 43.74 MPa

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.

Stiffnesses

k₁ = 200000·314.2/400 = 157080 N/mm

k₂ = 70000·706.9/500 = 98960 N/mm

Share of segment 2

F₂ = 80000·98960/256040 = 30920 N

Stress

σ₂ = 30920/706.9 = 43.74 MPa (compression)

Answer

σ₂ = 43.74 MPa

3.Torsion of shafts

Answer: φ = 2.882 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·800/(80000·79522) = 0.0503 rad

φ = 0.0503·180/π = 2.882°

Answer

φ = 2.882°

4.Torsion of shafts

Answer: τ_max = 83.37 MPa

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 75000/104.7 = 716.2 N·m

Polar second moment of area

J = π(40⁴ − 30⁴)/32 = 171806 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 716197·20/171806 = 83.37 MPa

Answer

τ_max = 83.37 MPa

5.Shear force & bending moment

Answer: M_max = 100.3 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

ΣM_A = 0: R_B·4 = 80·1.5 + 45·2.5

R_B = 232.5/4 = 58.13 kN

ΣF_y = 0: R_A = 125 − 58.13 = 66.88 kN

Moment under each load

Keep the part left of each load.

under P₁: M = R_A·1.5 = 66.88·1.5 = 100.3 kN·m

under P₂: M = R_B·1.5 = 58.13·1.5 = 87.19 kN·m

Largest value

M_max = 100.3 kN·m (under P₁)

Answer

M_max = 100.3 kN·m

6.Shear force & bending moment

Answer: M_x = 26.25 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·5 = 50 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 60·2.5 + 10·5·2.5

R_B = 275/5 = 55 kN

ΣF_y = 0: R_A = 110 − 55 = 55 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 = 55·4.5 − 60·(4.5 − 2.5) − 10·4.5·2.25 = 26.25 kN·m

Answer

M_x = 26.25 kN·m

7.Centroid & second moment of area

Answer: I_x = 35.42 × 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 two rectangles

Measure every y from the bottom edge.

flange: A_f = 120·15 = 1800 mm², y_f = 220 + 15/2 = 227.5 mm

web: A_w = 20·220 = 4400 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1800·227.5 + 4400·110)/6200 = 144.1 mm

Parallel-axis theorem

flange: 120·15³/12 + 1800·(227.5 − 144.1)² = 1.255 × 10⁷ mm⁴

web: 20·220³/12 + 4400·(110 − 144.1)² = 2.287 × 10⁷ mm⁴

I_x = 1.255 × 10⁷ + 2.287 × 10⁷ = 3.542 × 10⁷ mm⁴ = 35.42 × 10⁶ mm⁴

Answer

I_x = 35.42 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 8.293 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 180·30 = 5400 mm², y_f = 200 + 30/2 = 215 mm

web: A_w = 12·200 = 2400 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (5400·215 + 2400·100)/7800 = 179.6 mm

Parallel-axis theorem

flange: 180·30³/12 + 5400·(215 − 179.6)² = 7.166 × 10⁶ mm⁴

web: 12·200³/12 + 2400·(100 − 179.6)² = 2.321 × 10⁷ mm⁴

I_x = 7.166 × 10⁶ + 2.321 × 10⁷ = 3.038 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 230 − 179.6 = 50.38 mm

σ = M·y/I = 5 × 10⁶·50.38/3.038 × 10⁷ = 8.293 MPa

Answer

σ_top = 8.293 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 43.15 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′ = 95 − (−15)·(−0.1736) − 50·0.9848 = 43.15 MPa

Check

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

σ_x′ = 146.8 MPa, σ_x′ + σ_y′ = 190 = σ_x + σ_y

Answer

σ_y′ = 43.15 MPa

10.Mohr’s circle & stress transformation

Answer: τ_θ = 53.48 MPa

Transformation equation

τ_θ = −−55·0.5 + 30·0.866 = 53.48 MPa

Answer

τ_θ = 53.48 MPa

11.Beam deflection

Answer: δ = 2.917 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 = 29000 N, w = 8 N/mm, L = 2000 mm

EI = 3.2 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(3EI) = 29000·2000³/(3·3.2 × 10¹³) = 2.417 mm

δ_w = wL⁴/(8EI) = 8·2000⁴/(8·3.2 × 10¹³) = 0.5 mm

Add them

δ = 2.417 + 0.5 = 2.917 mm

Answer

δ = 2.917 mm

12.Beam deflection

Answer: R_B = 28.13 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·15·5/8 = 28.13 kN

Answer

R_B = 28.13 kN

13.Column buckling

Answer: P_cr = 62.01 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·20⁴/64 = 7854 mm⁴

Effective length

KL = 0.5·1000 = 500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·7854/500² = 62013 N = 62.01 kN

Answer

P_cr = 62.01 kN

14.Column buckling

Answer: d = 56.39 mm

Required critical load

P_cr = n·P = 2·40000 = 80000 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·80000·3500²/(π³·200000))^(1/4) = 56.39 mm

Answer

d = 56.39 mm

Check: the slenderness KL/r is 248.3, so the column is slender and Euler applies.

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