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 bar is made of a steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 500 mm, E = 70 GPaδ = 0.25 mm

P = kN

2.Axial stress & strainHard

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

P = 75 kNsegment 1: Ø25 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 250 mm rotate 1.17° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 50 N·md = 20 mmL = 250 mmφ = 1.17°

G = GPa

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1450 rpmd_o = 80 mmd_i = 48 mm

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 1.5 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 = 2.5 mP = 70 kN at 1.5 mw = 4 kN/m from 0 to 2.5 m

M_A = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 10 kN at 7 mw = 3 kN/m from 0 to 8 mx = 7 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 = 140 mmt_f = 30 mmt_w = 16 mmh_w = 280 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 100 mm × 200 mm plate has a rectangular cut-out 50 mm wide and 40 mm high, centred on the vertical centre line with its centre 120 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 200 mmcut-out: 50 × 40 mmhole centre: y = 120 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = 110 MPa and τₓᵧ = −30 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 80 MPaσ_y = 110 MPaτ_xy = −30 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 30 mm carries a bending moment of 200 N·m together with a torque of 1500 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τ_max.

d = 30 mmM = 200 N·mT = 1500 N·m

τ_max = 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 = 15 kN/mL = 3 mE = 11 GPa (timber)I = 23 × 10⁶ mm⁴

δ_max = 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 = 3 mw = 5 kN/m

R_B = kN

13.Column bucklingMedium

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

L = 2 mE = 70 GPa (aluminium)b × h = 50 × 100 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 2.5 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 28.94 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

A₁ = 490.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(70000·2827) = 2.526 × 10⁻⁶ mm/N

Load

P = 0.25/8.638 × 10⁻⁶ = 28942 N = 28.94 kN

Answer

P = 28.94 kN

2.Axial stress & strain

Answer: δ = 7.621 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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 152.8 MPa

σ₂ = 373 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 152.8·1500/100000 = 2.292 mm

δ₂ = σ₂L₂/E₂ = 373·1000/70000 = 5.329 mm

δ = δ₁ + δ₂ = 7.621 mm

Answer

δ = 7.621 mm

3.Torsion of shafts

Answer: G = 38.97 GPa

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 moment and angle

J = π·20⁴/32 = 15708 mm⁴

φ = 1.17·π/180 = 0.02042 rad

Shear modulus

T = 50 N·m = 50000 N·mm

G = TL/(Jφ) = 50000·250/(15708·0.02042) = 38970 MPa

G = 38.97 GPa

Answer

G = 38.97 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: τ_max = 7.526 MPa

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 100000/151.8 = 658.6 N·m

Polar second moment of area

J = π(80⁴ − 48⁴)/32 = 3.5 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 658572·40/3.5 × 10⁶ = 7.526 MPa

Answer

τ_max = 7.526 MPa

5.Shear force & bending moment

Answer: M_A = −117.5 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.

Keep the free-end part

P: −70·1.5 = −105 kN·m

w: −4·2.5²/2 = −12.5 kN·m

M_A = −105 − 12.5 = −117.5 kN·m

Answer

M_A = −117.5 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_x = 19.25 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·8 = 24 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 10·7 + 3·8·4

R_B = 166/8 = 20.75 kN

ΣF_y = 0: R_A = 34 − 20.75 = 13.25 kN

Cut at x = 7 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 = 13.25·7 − 3·7·3.5 = 19.25 kN·m

Answer

M_x = 19.25 kN·m

7.Centroid & second moment of area

Answer: I_x = 81.66 × 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 = 140·30 = 4200 mm², y_f = 280 + 30/2 = 295 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) = (4200·295 + 4480·140)/8680 = 215 mm

Parallel-axis theorem

flange: 140·30³/12 + 4200·(295 − 215)² = 2.72 × 10⁷ mm⁴

web: 16·280³/12 + 4480·(140 − 215)² = 5.447 × 10⁷ mm⁴

I_x = 2.72 × 10⁷ + 5.447 × 10⁷ = 8.166 × 10⁷ mm⁴ = 81.66 × 10⁶ mm⁴

Answer

I_x = 81.66 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 65.51 × 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 = (20000·100 − 2000·120)/(20000 − 2000) = 97.78 mm

Full plate about ȳ

100·200³/12 + 20000·2.222² = 66.77 × 10⁶ mm⁴

Hole about ȳ

50·40³/12 + 2000·22.22² = 1.254 × 10⁶ mm⁴

Subtract

I_x = 65.51 × 10⁶ mm⁴

Answer

I_x = 65.51 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_abs = 64.27 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 95 MPa, R = √((−15)² + (−30)²) = 33.54 MPa

σ₁ = 128.5 MPa, σ₂ = 61.46 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(67.08, 128.5, 61.46)/2 = 64.27 MPa

Answer

τ_abs = 64.27 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: τ_max = 285.4 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·200000/(π·30³) = 75.45 MPa

τ = 16T/(πd³) = 16·1500000/(π·30³) = 282.9 MPa

Mohr’s circle

C = σ/2 = 37.73 MPa

R = √(37.73² + 282.9²) = 285.4 MPa

Answer

τ_max = 285.4 MPa

11.Beam deflection

Answer: δ_max = 62.53 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 = 15 N/mm, L = 3000 mm

EI = 2.53 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·15·3000⁴/(384·2.53 × 10¹¹) = 62.53 mm

Answer

δ_max = 62.53 mm

12.Beam deflection

Answer: R_B = 5.625 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 = 3·5·3/8 = 5.625 kN

Answer

R_B = 5.625 kN

13.Column buckling

Answer: P_cr = 367.2 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 100·50³/12 = 1.042 × 10⁶ mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·1.042 × 10⁶/1400² = 367173 N = 367.2 kN

Answer

P_cr = 367.2 kN

14.Column buckling

Answer: σ_cr = 36.26 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 = 0.7·2500 = 1750 mm

KL/r = 1750/7.5 = 233.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/233.3² = 36.26 MPa

Answer

σ_cr = 36.26 MPa

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