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 elongation δ of the copper bar under the axial load.

P = 10 kNd = 20 mmL = 1.5 mE = 117 GPa

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 = 110 kNsegment 1: Ø35 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 1 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid brass 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 = 25 mmL = 0.8 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 50 kW at 500 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter d_o.

P = 50 kWN = 500 rpmd_i/d_o = 0.5τ_allow = 80 MPa

d_o = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 35 kN·m at x = 1 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 6 mM₀ = 35 kN·m clockwise, at 1 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 1 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance x_c from A to the point of contraflexure between the supports (where the bending moment changes sign).

L = 5 mw = 3 kN/m from 0 to 5 msupports at x = 0 and x = 4 m

x_c = 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 = 300 mmt_f = 20 mmt_w = 18 mmh_w = 140 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 200 mm plate has a rectangular cut-out 50 mm wide and 30 mm high, centred on the vertical centre line with its centre 130 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 200 mmcut-out: 50 × 30 mmhole centre: y = 130 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 90 MPa, σᵧ = 20 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 90 MPaσ_y = 20 MPaτ_xy = 30 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 5 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1000 mmt = 5 mmp = 2.5 MPaα = 20°

σ_n = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 4 kN/mL = 6 mE = 200 GPa (steel)I = 84 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 20 kN/mL = 3 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 6 mE = 200 GPa (steel)d = 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 90 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 90 kNn = 2L = 6 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.4081 mm

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·20²/4 = 314.2 mm²

σ = P/A = 10·1000/314.2 = 31.83 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 31.83/117000 = 0.0002721 = 272.1 µε

Elongation

δ = ε·L = 0.0002721·1500 mm = 0.4081 mm

Answer

δ = 0.4081 mm

2.Axial stress & strain

Answer: δ = 7.126 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·35²/4 = 962.1 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 114.3 MPa

σ₂ = 547.1 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 114.3·1500/70000 = 2.45 mm

δ₂ = σ₂L₂/E₂ = 547.1·1000/117000 = 4.676 mm

δ = δ₁ + δ₂ = 7.126 mm

Answer

δ = 7.126 mm

3.Torsion of shafts

Answer: φ = 12.26 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·800/(39000·38350) = 0.214 rad

φ = 0.214·180/π = 12.26°

Answer

φ = 12.26°

4.Torsion of shafts

Answer: d_o = 40.18 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 50000/52.36 = 954.9 N·m

Outer diameter from the allowable stress

T = 954.9 N·m = 954930 N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·954930/(π·80·0.9375)) = 40.18 mm

d_i = 0.5·40.18 = 20.09 mm

Answer

d_o = 40.18 mm

A solid shaft would need d = 39.32 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: M_x = −5.833 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 35/6 = 5.833 kN (upward)

R_A = −R_B = −5.833 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −5.833·1 = −5.833 kN·m

Answer

M_x = −5.833 kN·m

6.Shear force & bending moment

Answer: x_c = 3.75 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 = 3·5 = 15 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·4 = 3·5·2.5

R_B = 37.5/4 = 9.375 kN

ΣF_y = 0: R_A = 15 − 9.375 = 5.625 kN

Zero moment

Between the supports M(x) = R_A·x − w·x²/2. Apart from x = 0 it is zero at x = 2R_A/w.

R_A·x_c − w·x_c²/2 = 0

x_c = 2·5.625/3 = 3.75 m

Answer

x_c = 3.75 m

7.Centroid & second moment of area

Answer: I_x = 15.67 × 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 = 300·20 = 6000 mm², y_f = 140 + 20/2 = 150 mm

web: A_w = 18·140 = 2520 mm², y_w = 140/2 = 70 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·150 + 2520·70)/8520 = 126.3 mm

Parallel-axis theorem

flange: 300·20³/12 + 6000·(150 − 126.3)² = 3.559 × 10⁶ mm⁴

web: 18·140³/12 + 2520·(70 − 126.3)² = 1.211 × 10⁷ mm⁴

I_x = 3.559 × 10⁶ + 1.211 × 10⁷ = 1.567 × 10⁷ mm⁴ = 15.67 × 10⁶ mm⁴

Answer

I_x = 15.67 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 78.45 × 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 = (24000·100 − 1500·130)/(24000 − 1500) = 98 mm

Full plate about ȳ

120·200³/12 + 24000·2² = 80.1 × 10⁶ mm⁴

Hole about ȳ

50·30³/12 + 1500·32² = 1.649 × 10⁶ mm⁴

Subtract

I_x = 78.45 × 10⁶ mm⁴

Answer

I_x = 78.45 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 46.1 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.

Radius of Mohr’s circle

τ_max = R = √(((90 − 20)/2)² + 30²) = 46.1 MPa

Answer

τ_max = 46.1 MPa

10.Mohr’s circle & stress transformation

Answer: σ_n = 235.4 MPa

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 2.5·500/5 = 250 MPa

σ_a = pr/(2t) = 125 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 125·sin²20° + 250·cos²20° = 235.4 MPa

Answer

σ_n = 235.4 MPa

11.Beam deflection

Answer: δ = 38.57 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 = 4 N/mm, L = 6000 mm

EI = 1.68 × 10¹³ N·mm²

Deflection

δ = wL⁴/(8EI) = 4·6000⁴/(8·1.68 × 10¹³) = 38.57 mm

Answer

δ = 38.57 mm

12.Beam deflection

Answer: I_req = 12.66 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 3000/360 = 8.333 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·20·3000⁴/(384·200000·8.333) = 1.266 × 10⁷ mm⁴

I_req = 12.66 × 10⁶ mm⁴

Answer

I_req = 12.66 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 6.89 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·125664/6000² = 6890 N = 6.89 kN

Answer

P_cr = 6.89 kN

14.Column buckling

Answer: d = 166.3 mm

Required critical load

P_cr = n·P = 2·90000 = 180000 N

KL = 2·6000 = 12000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·180000·12000²/(π³·70000))^(1/4) = 166.3 mm

Answer

d = 166.3 mm

Check: the slenderness KL/r is 288.7, so the column is slender and Euler applies.

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