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 steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 50 mm, E = 120 GPa). Both are 800 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.

P = 200 kNrod: Ø20 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 120 GPaL = 800 mm

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 45 kNsegment 1: Ø30 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 1.5 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 40 mm and length 0.5 m; segment BC has diameter 20 mm and length 0.5 m. A torque of 200 N·m acts at the free end C. Find the maximum shear stress τ_max in the shaft.

T = 200 N·md₁ = 40 mm, L₁ = 0.5 md₂ = 20 mm, L₂ = 0.5 m

τ_max = MPa

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 1000 N·m is applied at C, 2 m from A and 1.2 m from B. Find the maximum shear stress τ_max in the shaft.

d = 30 mmT₀ = 1000 N·ma = 2 mb = 1.2 m

τ_max = MPa

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 8 kN/m from 0 to 5 mx = 3 m

V_x = kN

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 7.5 mw = 2 kN/m from 0 to 7.5 msupports at x = 0 and x = 5 m

M_max = 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 = 260 mmt_f = 15 mmt_w = 20 mmh_w = 200 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 20 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 200 mmt_f = 10 mmt_w = 8 mmh_w = 100 mmM = 20 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −80 MPa, σᵧ = −40 MPa and τₓᵧ = −30 MPa (tension positive). The element is rotated 25° 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 = −40 MPaτ_xy = −30 MPaθ = 25°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 0 MPa, σᵧ = −20 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.

σ_x = 0 MPaσ_y = −20 MPaτ_xy = −50 MPaθ = 30°

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 = 2 mE = 11 GPa (timber)I = 12 × 10⁶ mm⁴

θ_A = rad

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 2 m from support B. Find the maximum deflection δ_max of the beam.

P = 16 kNb = 2 mL = 5 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 2.5 mE = 200 GPa (steel)b × h = 50 × 60 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 135 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 135 kNn = 2L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.3717 mm

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·20²/4 = 314.2 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 120·10³·3063 = 4.304 × 10⁸ N

Shortening

δ = 200000·800/4.304 × 10⁸ = 0.3717 mm

Answer

δ = 0.3717 mm

2.Axial stress & strain

Answer: δ = 3.029 mm

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·30²/4 = 706.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 63.66 MPa

σ₂ = 223.8 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 63.66·500/200000 = 0.1592 mm

δ₂ = σ₂L₂/E₂ = 223.8·1500/117000 = 2.869 mm

δ = δ₁ + δ₂ = 3.029 mm

Answer

δ = 3.029 mm

3.Torsion of shafts

Answer: τ_max = 127.3 MPa

Stress in each segment

T = 200 N·m = 200000 N·mm

τ_AB = 16·200000/(π·40³) = 15.92 MPa

τ_BC = 16·200000/(π·20³) = 127.3 MPa

Answer

τ_max = 127.3 MPa

4.Torsion of shafts

Answer: τ_max = 117.9 MPa

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. When both ends are fixed, statics alone gives only T_A + T_B = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

T_A + T_B = 1000

T_A·a = T_B·b

T_A = 1000·1.2/3.2 = 375 N·m

T_B = 1000·2/3.2 = 625 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τ_max = 16·625000/(π·30³) = 117.9 MPa

Answer

τ_max = 117.9 MPa

5.Shear force & bending moment

Answer: V_x = −4 kN

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·5 = 40 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 8·5·2.5

R_B = 100/5 = 20 kN

ΣF_y = 0: R_A = 40 − 20 = 20 kN

Cut at x = 3 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.

V = R_A − (all downward loads left of the cut)

V_x = 20 − 8·3 = −4 kN

Answer

V_x = −4 kN

6.Shear force & bending moment

Answer: M_max = 3.516 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 2·7.5 = 15 kN, acting at x̄ = 3.75 m

ΣM_A = 0: R_B·5 = 2·7.5·3.75

R_B = 56.25/5 = 11.25 kN

ΣF_y = 0: R_A = 15 − 11.25 = 3.75 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 3.75/2 = 1.875 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 3.75²/(2·2) = 3.516 kN·m

Answer

M_max = 3.516 kN·m

7.Centroid & second moment of area

Answer: I_x = 36.23 × 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 = 260·15 = 3900 mm², y_f = 200 + 15/2 = 207.5 mm

web: A_w = 20·200 = 4000 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3900·207.5 + 4000·100)/7900 = 153.1 mm

Parallel-axis theorem

flange: 260·15³/12 + 3900·(207.5 − 153.1)² = 1.163 × 10⁷ mm⁴

web: 20·200³/12 + 4000·(100 − 153.1)² = 2.46 × 10⁷ mm⁴

I_x = 1.163 × 10⁷ + 2.46 × 10⁷ = 3.623 × 10⁷ mm⁴ = 36.23 × 10⁶ mm⁴

Answer

I_x = 36.23 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 171.8 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 200·10 = 2000 mm², y_f = 100 + 10/2 = 105 mm

web: A_w = 8·100 = 800 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2000·105 + 800·50)/2800 = 89.29 mm

Parallel-axis theorem

flange: 200·10³/12 + 2000·(105 − 89.29)² = 510544 mm⁴

web: 8·100³/12 + 800·(50 − 89.29)² = 1.901 × 10⁶ mm⁴

I_x = 510544 + 1.901 × 10⁶ = 2.412 × 10⁶ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 110 − 89.29 = 20.71 mm

σ = M·y/I = 2 × 10⁷·20.71/2.412 × 10⁶ = 171.8 MPa

Answer

σ_top = 171.8 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = −24.16 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′ = −60 − (−20)·0.6428 − (−30)·0.766 = −24.16 MPa

Check

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

σ_x′ = −95.84 MPa, σ_x′ + σ_y′ = −120 = σ_x + σ_y

Answer

σ_y′ = −24.16 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −48.3 MPa

Transformation equation

σ_θ = −10 + 10·0.5 + −50·0.866 = −48.3 MPa

Answer

σ_θ = −48.3 MPa

11.Beam deflection

Answer: θ_A = 0.03788 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 = 2000 mm

EI = 1.32 × 10¹¹ N·mm²

End slope

θ_A = wL³/(24EI) = 15·2000³/(24·1.32 × 10¹¹) = 0.03788 rad

Answer

θ_A = 0.03788 rad

That is 2.17°.

12.Beam deflection

Answer: δ_max = 63.01 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 = 16000 N, b = 2000 mm, L = 5000 mm

EI = 6.27 × 10¹¹ N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((5000² − 2000²)/3) = 2646 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 63.01 mm

Answer

δ_max = 63.01 mm

13.Column buckling

Answer: P_cr = 402.8 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 = 60·50³/12 = 625000 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·625000/1750² = 402841 N = 402.8 kN

Answer

P_cr = 402.8 kN

14.Column buckling

Answer: d = 59.21 mm

Required critical load

P_cr = n·P = 2·135000 = 270000 N

KL = 0.7·3000 = 2100 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·270000·2100²/(π³·200000))^(1/4) = 59.21 mm

Answer

d = 59.21 mm

Check: the slenderness KL/r is 141.9, so the column is slender and Euler applies.

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