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 steel bar under the axial load.

P = 120 kNd = 10 mmL = 2.5 mE = 200 GPa

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 60 mm, E = 100 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø100/Ø60 mm, E = 100 GPa, σ_allow = 50 MPa

P_max = kN

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 = 500 N·md = 40 mmL = 2.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 25 mm is fixed at A. A torque of 500 N·m acts at B, 0.8 m from A, and a torque of 100 N·m in the same direction acts at the free end C, a further 1.5 m along. Find the maximum shear stress τ_max in the shaft.

d = 25 mmT_B = 500 N·mT_C = 100 N·mAB = 0.8 mBC = 1.5 m

τ_max = MPa

5.Shear force & bending momentMedium

The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 2 m from A. Find the bending moment M_A at the fixed support A. Use sagging moments positive and upward shear on the left face positive.

L = 2 mP = 30 kN at 2 mw = 5 kN/m from 0 to 2 m

M_A = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2 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 = 10 mw = 6 kN/m from 0 to 10 msupports at x = 0 and x = 8 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 90 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 150 mmd = 30 mmhole centre: y = 90 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 25 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 240 mmt_f = 10 mmt_w = 16 mmh_w = 280 mmM = 25 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 110 MPa, σᵧ = −60 MPa and τₓᵧ = −10 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 = 110 MPaσ_y = −60 MPaτ_xy = −10 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −30 MPa, σᵧ = 30 MPa and τₓᵧ = 50 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −30 MPaσ_y = 30 MPaτ_xy = 50 MPaσ_Y = 350 MPa

n =

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 = 4 kNw = 4 kN/mL = 2 mE = 11 GPa (timber)I = 120 × 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 = 5 kN/mL = 4 mE = 70 GPa (aluminium)

I_req = × 10⁶ mm⁴

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 = 4 mE = 200 GPa (steel)b × h = 40 × 100 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 = 1.5 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 19.1 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·10²/4 = 78.54 mm²

σ = P/A = 120·1000/78.54 = 1528 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 1528/200000 = 0.007639 = 7639 µε

Elongation

δ = ε·L = 0.007639·2500 mm = 19.1 mm

Answer

δ = 19.1 mm

2.Axial stress & strain

Answer: P_max = 282.7 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(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·5027 = 5.655 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0005) = 0.0005 (brass)

Largest load

P_max = 0.0005·5.655 × 10⁸ = 282743 N = 282.7 kN

Answer

P_max = 282.7 kN

3.Torsion of shafts

Answer: φ = 7.307 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·2500/(39000·251327) = 0.1275 rad

φ = 0.1275·180/π = 7.307°

Answer

φ = 7.307°

4.Torsion of shafts

Answer: τ_max = 195.6 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 100 N·m

T_AB = T_C + T_B = 100 + 500 = 600 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 600 N·m

τ_max = 16·600000/(π·25³) = 195.6 MPa

Answer

τ_max = 195.6 MPa

5.Shear force & bending moment

Answer: M_A = −70 kN·m

Keep the free-end part

P: −30·2 = −60 kN·m

w: −5·2²/2 = −10 kN·m

M_A = −60 − 10 = −70 kN·m

Answer

M_A = −70 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_max = 42.19 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 = 6·10 = 60 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·8 = 6·10·5

R_B = 300/8 = 37.5 kN

ΣF_y = 0: R_A = 60 − 37.5 = 22.5 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 22.5/6 = 3.75 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 22.5²/(2·6) = 42.19 kN·m

Answer

M_max = 42.19 kN·m

7.Centroid & second moment of area

Answer: I_x = 27.92 × 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 = (15000·75 − 706.9·90)/(15000 − 706.9) = 74.26 mm

Full plate about ȳ

100·150³/12 + 15000·0.7418² = 28.13 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·15.74² = 0.2149 × 10⁶ mm⁴

Subtract

I_x = 27.92 × 10⁶ mm⁴

Answer

I_x = 27.92 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 76.67 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 = 240·10 = 2400 mm², y_f = 280 + 10/2 = 285 mm

web: A_w = 16·280 = 4480 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2400·285 + 4480·140)/6880 = 190.6 mm

Parallel-axis theorem

flange: 240·10³/12 + 2400·(285 − 190.6)² = 2.142 × 10⁷ mm⁴

web: 16·280³/12 + 4480·(140 − 190.6)² = 4.073 × 10⁷ mm⁴

I_x = 2.142 × 10⁷ + 4.073 × 10⁷ = 6.215 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 190.6 mm

σ = M·y/I = 2.5 × 10⁷·190.6/6.215 × 10⁷ = 76.67 MPa

Answer

σ_bottom = 76.67 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −3.355 °

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·−10, 110 − −60) = −6.71°

Angle on the element

θ_p = −3.355°

Answer

θ_p = −3.355°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 3.466

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 0 MPa, R = 58.31 MPa

σ₁ = 58.31 MPa, σ₂ = −58.31 MPa, σ₃ = 0

Von Mises

σ_vM = √(58.31² − 58.31·(−58.31) + (−58.31)²) = 101 MPa

n = 350/101 = 3.466

Answer

n = 3.466

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ = 14.14 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 = 4000 N, w = 4 N/mm, L = 2000 mm

EI = 1.32 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 4000·2000³/(3·1.32 × 10¹²) = 8.081 mm

δ_w = wL⁴/(8EI) = 4·2000⁴/(8·1.32 × 10¹²) = 6.061 mm

Add them

δ = 8.081 + 6.061 = 14.14 mm

Answer

δ = 14.14 mm

12.Beam deflection

Answer: I_req = 21.43 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·5·4000⁴/(384·70000·11.11) = 2.143 × 10⁷ mm⁴

I_req = 21.43 × 10⁶ mm⁴

Answer

I_req = 21.43 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 16.45 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·533333/8000² = 16449 N = 16.45 kN

Answer

P_cr = 16.45 kN

14.Column buckling

Answer: σ_cr = 197.4 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 = 0.5·1500 = 750 mm

KL/r = 750/7.5 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/100² = 197.4 MPa

Answer

σ_cr = 197.4 MPa

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