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 = 40 kNd = 12 mmL = 0.5 mE = 117 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 200 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 200 mm, E = 70 GPaP = 100 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid aluminium 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 = 0.8 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 15 kWN = 500 rpmτ_allow = 80 MPa

d = mm

5.Shear force & bending momentMedium

The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.

45 kN @ 3 m10 kN/m, 0–6 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mP = 45 kN at 5 mw = 12 kN/m from 0 to 6 mx = 5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 150 mm × 240 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 72 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 240 mmd = 60 mmhole centre: y = 72 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 120 mm × 240 mm plate has a rectangular cut-out 50 mm wide and 30 mm high, centred on the vertical centre line with its centre 84 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 240 mmcut-out: 50 × 30 mmhole centre: y = 84 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −20 MPa, σᵧ = 40 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −20 MPaσ_y = 40 MPaτ_xy = −40 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 30 mm carries a bending moment of 1000 N·m together with a torque of 1000 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 = 1000 N·mT = 1000 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 = 10 kN/mL = 3 mE = 70 GPa (aluminium)I = 8.5 × 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 = 8 mw = 4 kN/m

R_B = kN

13.Column bucklingMedium

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

L = 3 mE = 70 GPa (aluminium)d = 30 mm

P_cr = kN

14.Column bucklingHard

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

L = 2.5 mE = 200 GPa (steel)d = 25 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 1.511 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·12²/4 = 113.1 mm²

σ = P/A = 40·1000/113.1 = 353.7 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 353.7/117000 = 0.003023 = 3023 µε

Elongation

δ = ε·L = 0.003023·500 mm = 1.511 mm

Answer

δ = 1.511 mm

2.Axial stress & strain

Answer: σ₂ = 48.7 MPa

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.

Stiffnesses

k₁ = 200000·706.9/300 = 471239 N/mm

k₂ = 70000·706.9/200 = 247400 N/mm

Share of segment 2

F₂ = 100000·247400/718639 = 34426 N

Stress

σ₂ = 34426/706.9 = 48.7 MPa (compression)

Answer

σ₂ = 48.7 MPa

3.Torsion of shafts

Answer: φ = 3.507 °

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) = 500·1000·800/(26000·251327) = 0.06121 rad

φ = 0.06121·180/π = 3.507°

Answer

φ = 3.507°

4.Torsion of shafts

Answer: d = 26.32 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 15000/52.36 = 286.5 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·286479/(π·80)) = 26.32 mm

Answer

d = 26.32 mm

In practice you would round up to the next standard size.

5.Shear force & bending moment

Answer: D (x = 3 m)

Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 52.5 kN, R₂ = 52.5 kN

Bending moment at every mark

Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.

M_A = 0, M_B = 47.5, M_C = 85, M_D = 112.5 kN·m, M_E = 85, M_F = 47.5, M_G = 0

Answer

Mark D: M_D = 112.5 kN·m

6.Shear force & bending moment

Answer: M_x = 67.5 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·6 = 72 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 45·5 + 12·6·3

R_B = 441/6 = 73.5 kN

ΣF_y = 0: R_A = 117 − 73.5 = 43.5 kN

Cut at x = 5 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 = 43.5·5 − 12·5·2.5 = 67.5 kN·m

Answer

M_x = 67.5 kN·m

7.Centroid & second moment of area

Answer: I_x = 165.1 × 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 = (36000·120 − 2827·72)/(36000 − 2827) = 124.1 mm

Full plate about ȳ

150·240³/12 + 36000·−4.091² = 173.4 × 10⁶ mm⁴

Hole about ȳ

π·60⁴/64 + 2827·−52.09² = 8.308 × 10⁶ mm⁴

Subtract

I_x = 165.1 × 10⁶ mm⁴

Answer

I_x = 165.1 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 136.1 × 10⁶ mm⁴

Centroid

ȳ = (28800·120 − 1500·84)/(28800 − 1500) = 122 mm

Full plate about ȳ

120·240³/12 + 28800·−1.978² = 138.4 × 10⁶ mm⁴

Hole about ȳ

50·30³/12 + 1500·−37.98² = 2.276 × 10⁶ mm⁴

Subtract

I_x = 136.1 × 10⁶ mm⁴

Answer

I_x = 136.1 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_max = 50 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 = √(((−20 − 40)/2)² + −40²) = 50 MPa

Answer

τ_max = 50 MPa

10.Mohr’s circle & stress transformation

Answer: τ_max = 266.8 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·1000000/(π·30³) = 377.3 MPa

τ = 16T/(πd³) = 16·1000000/(π·30³) = 188.6 MPa

Mohr’s circle

C = σ/2 = 188.6 MPa

R = √(188.6² + 188.6²) = 266.8 MPa

Answer

τ_max = 266.8 MPa

11.Beam deflection

Answer: δ_max = 17.73 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 = 10 N/mm, L = 3000 mm

EI = 5.95 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·10·3000⁴/(384·5.95 × 10¹¹) = 17.73 mm

Answer

δ_max = 17.73 mm

12.Beam deflection

Answer: R_B = 12 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 = 3·4·8/8 = 12 kN

Answer

R_B = 12 kN

13.Column buckling

Answer: P_cr = 12.21 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·39761/1500² = 12209 N = 12.21 kN

Answer

P_cr = 12.21 kN

14.Column buckling

Answer: σ_cr = 12.34 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 1·2500 = 2500 mm

KL/r = 2500/6.25 = 400

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/400² = 12.34 MPa

Answer

σ_cr = 12.34 MPa

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