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

P = 80 kNd = 16 mmL = 1 mE = 117 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø120/Ø50 mm, E = 100 GPa, σ_allow = 100 MPa

P_max = kN

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 = 50 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 75 kW at 1450 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

P = 75 kWN = 1450 rpmd_i/d_o = 0.5τ_allow = 50 MPa

d_o = mm

5.Shear force & bending momentMedium

Find the bending moment M_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 = 5 kN/m from 0 to 5 mx = 3 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 9 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 25 kN at 9 mw = 8 kN/m from 0 to 10 mx = 9 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 = 260 mmt_f = 15 mmt_w = 8 mmh_w = 280 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 bending stress σ_bottom at the bottom edge.

b_f = 160 mmt_f = 20 mmt_w = 10 mmh_w = 240 mmM = 20 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −40 MPa, σᵧ = 70 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 = −40 MPaσ_y = 70 MPaτ_xy = −50 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 30 MPa, σᵧ = 50 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 30 MPaσ_y = 50 MPaτ_xy = 40 MPaθ = -45°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum deflection δ_max, at midspan.

w = 5 kN/mL = 4 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 12 kNw = 10 kN/mL = 6 mE = 70 GPa (aluminium)I = 160 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 3 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 75 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 75 kNn = 2.5L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 3401 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 = 80·1000/201.1 = 397.9 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 397.9/117000 = 0.003401 = 3401 µε

Answer

ε = 3401 µε

2.Axial stress & strain

Answer: P_max = 997.5 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

A_r = π·20²/4 = 314.2 mm²

A_t = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·9346 = 9.975 × 10⁸ N

Governing strain

ε = min(0.001, 0.001) = 0.001 (brass)

Largest load

P_max = 0.001·9.975 × 10⁸ = 997456 N = 997.5 kN

Answer

P_max = 997.5 kN

3.Torsion of shafts

Answer: φ = 0.7003 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·1500/(80000·613592) = 0.01222 rad

φ = 0.01222·180/π = 0.7003°

Answer

φ = 0.7003°

4.Torsion of shafts

Answer: d_o = 37.72 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 75000/151.8 = 493.9 N·m

Outer diameter from the allowable stress

T = 493.9 N·m = 493929 N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·493929/(π·50·0.9375)) = 37.72 mm

d_i = 0.5·37.72 = 18.86 mm

Answer

d_o = 37.72 mm

A solid shaft would need d = 36.92 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: M_x = 15 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 = 5·5 = 25 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 5·5·2.5

R_B = 62.5/5 = 12.5 kN

ΣF_y = 0: R_A = 25 − 12.5 = 12.5 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.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 12.5·3 − 5·3·1.5 = 15 kN·m

Answer

M_x = 15 kN·m

6.Shear force & bending moment

Answer: M_x = 58.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·10 = 80 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 25·9 + 8·10·5

R_B = 625/10 = 62.5 kN

ΣF_y = 0: R_A = 105 − 62.5 = 42.5 kN

Cut at x = 9 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 = 42.5·9 − 8·9·4.5 = 58.5 kN·m

Answer

M_x = 58.5 kN·m

7.Centroid & second moment of area

Answer: I_x = 45.66 × 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 = 280 + 15/2 = 287.5 mm

web: A_w = 8·280 = 2240 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3900·287.5 + 2240·140)/6140 = 233.7 mm

Parallel-axis theorem

flange: 260·15³/12 + 3900·(287.5 − 233.7)² = 1.137 × 10⁷ mm⁴

web: 8·280³/12 + 2240·(140 − 233.7)² = 3.43 × 10⁷ mm⁴

I_x = 1.137 × 10⁷ + 3.43 × 10⁷ = 4.566 × 10⁷ mm⁴ = 45.66 × 10⁶ mm⁴

Answer

I_x = 45.66 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 111.6 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 160·20 = 3200 mm², y_f = 240 + 20/2 = 250 mm

web: A_w = 10·240 = 2400 mm², y_w = 240/2 = 120 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3200·250 + 2400·120)/5600 = 194.3 mm

Parallel-axis theorem

flange: 160·20³/12 + 3200·(250 − 194.3)² = 1.004 × 10⁷ mm⁴

web: 10·240³/12 + 2400·(120 − 194.3)² = 2.476 × 10⁷ mm⁴

I_x = 1.004 × 10⁷ + 2.476 × 10⁷ = 3.48 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 194.3 mm

σ = M·y/I = 2 × 10⁷·194.3/3.48 × 10⁷ = 111.6 MPa

Answer

σ_bottom = 111.6 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −68.86 °

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·−50, −40 − 70) = −137.7°

Angle on the element

θ_p = −68.86°

Answer

θ_p = −68.86°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = −10 MPa

Transformation equation

τ_θ = −−10·−1 + 40·0 = −10 MPa

Answer

τ_θ = −10 MPa

11.Beam deflection

Answer: δ_max = 2.315 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 = 5 N/mm, L = 4000 mm

EI = 7.2 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·5·4000⁴/(384·7.2 × 10¹²) = 2.315 mm

Answer

δ_max = 2.315 mm

12.Beam deflection

Answer: δ_max = 19.89 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 = 12000 N, w = 10 N/mm, L = 6000 mm, EI = 1.12 × 10¹³ N·mm²

Each load on its own

δ_P = PL³/(48EI) = 12000·6000³/(48·1.12 × 10¹³) = 4.821 mm

δ_w = 5wL⁴/(384EI) = 5·10·6000⁴/(384·1.12 × 10¹³) = 15.07 mm

Add them

δ_max = 4.821 + 15.07 = 19.89 mm

Answer

δ_max = 19.89 mm

13.Column buckling

Answer: P_cr = 110.2 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/6000² = 110245 N = 110.2 kN

Answer

P_cr = 110.2 kN

14.Column buckling

Answer: d = 64.6 mm

Required critical load

P_cr = n·P = 2.5·75000 = 187500 N

KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·187500·3000²/(π³·200000))^(1/4) = 64.6 mm

Answer

d = 64.6 mm

Check: the slenderness KL/r is 185.8, so the column is slender and Euler applies.

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