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

P = 90 kNd = 10 mmL = 1.5 mE = 70 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø80/Ø50 mm, E = 100 GPa, σ_allow = 70 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 = 750 N·md = 30 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 750 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 5 kWN = 750 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

Find the shear force V_x at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 3 kN/m from 0 to 8 mx = 4.5 m

V_x = kN

6.Shear force & bending momentHard

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.

30 kN @ 0 m15 kN @ 12 m55 kN @ 6 m8 kN/m, 2–10 m

1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)

7.Centroid & second moment of areaMedium

A 150 mm × 200 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 70 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 200 mmd = 60 mmhole centre: y = 70 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 40 MPa in tension and 100 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 100 mmt_f = 10 mmt_w = 14 mmh_w = 140 mmσ_t,allow = 40 MPaσ_c,allow = 100 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 120 MPa, σᵧ = 120 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 = 120 MPaσ_y = 120 MPaτ_xy = −30 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 140 MPa, σᵧ = −10 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° clockwise from the x-axis.

σ_x = 140 MPaσ_y = −10 MPaτ_xy = −60 MPaθ = -20°

σ_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 = 4 kN/mL = 5 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 2.5 m from support B. Find the maximum deflection δ_max of the beam.

P = 4 kNb = 2.5 mL = 8 mE = 200 GPa (steel)I = 57 × 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 = 5 mE = 200 GPa (steel)d = 80 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 45 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 45 kNn = 2.5L = 2 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 16370 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·10²/4 = 78.54 mm²

σ = P/A = 90·1000/78.54 = 1146 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 1146/70000 = 0.01637 = 16370 µε

Answer

ε = 16370 µε

2.Axial stress & strain

Answer: P_max = 313.4 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·30²/4 = 706.9 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 100·10³·3063 = 4.477 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0007) = 0.0007 (brass)

Largest load

P_max = 0.0007·4.477 × 10⁸ = 313374 N = 313.4 kN

Answer

P_max = 313.4 kN

3.Torsion of shafts

Answer: φ = 10.13 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1500/(80000·79522) = 0.1768 rad

φ = 0.1768·180/π = 10.13°

Answer

φ = 10.13°

4.Torsion of shafts

Answer: d = 20.09 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 5000/78.54 = 63.66 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·63662/(π·40)) = 20.09 mm

Answer

d = 20.09 mm

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

5.Shear force & bending moment

Answer: V_x = −1.5 kN

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 = 3·8 = 24 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 3·8·4

R_B = 96/8 = 12 kN

ΣF_y = 0: R_A = 24 − 12 = 12 kN

Cut at x = 4.5 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.

V = R_A − (all downward loads left of the cut)

V_x = 12 − 3·4.5 = −1.5 kN

Answer

V_x = −1.5 kN

6.Shear force & bending moment

Answer: D (x = 6 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 = 93.25 kN, R₂ = 70.75 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 = −60, M_C = 50.5, M_D = 129 kN·m, M_E = 65.5, M_F = −30, M_G = 0

Answer

Mark D: M_D = 129 kN·m

7.Centroid & second moment of area

Answer: I_x = 96.55 × 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 = (30000·100 − 2827·70)/(30000 − 2827) = 103.1 mm

Full plate about ȳ

150·200³/12 + 30000·−3.122² = 100.3 × 10⁶ mm⁴

Hole about ȳ

π·60⁴/64 + 2827·−33.12² = 3.738 × 10⁶ mm⁴

Subtract

I_x = 96.55 × 10⁶ mm⁴

Answer

I_x = 96.55 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 2.909 kN·m

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.

Centroid

ȳ = (1000·145 + 1960·70)/2960 = 95.34 mm

top edge: c = 150 − 95.34 = 54.66 mm

Second moment of area

I = 100·10³/12 + 1000·49.66² + 14·140³/12 + 1960·25.34² = 6.934 × 10⁶ mm⁴

Moment allowed by each edge

Sagging: the top is in compression, the bottom in tension.

top edge: M = 100·6.934 × 10⁶/54.66 = 12.69 kN·m

bottom edge: M = 40·6.934 × 10⁶/95.34 = 2.909 kN·m

Governing value

M_allow = min = 2.909 kN·m (bottom edge)

Answer

M_allow = 2.909 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_abs = 75 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 = 120 MPa, R = √(0² + (−30)²) = 30 MPa

σ₁ = 150 MPa, σ₂ = 90 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(60, 150, 90)/2 = 75 MPa

Answer

τ_abs = 75 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: σ_θ = 161 MPa

Transformation equation

σ_θ = 65 + 75·0.766 + −60·−0.6428 = 161 MPa

Answer

σ_θ = 161 MPa

11.Beam deflection

Answer: δ_max = 7.077 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 = 5000 mm

EI = 4.6 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·4·5000⁴/(384·4.6 × 10¹²) = 7.077 mm

Answer

δ_max = 7.077 mm

12.Beam deflection

Answer: δ_max = 3.087 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 = 4000 N, b = 2500 mm, L = 8000 mm

EI = 1.14 × 10¹³ N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((8000² − 2500²)/3) = 4387 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 3.087 mm

Answer

δ_max = 3.087 mm

13.Column buckling

Answer: P_cr = 39.69 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 = 2·5000 = 10000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/10000² = 39688 N = 39.69 kN

Answer

P_cr = 39.69 kN

14.Column buckling

Answer: d = 85.35 mm

Required critical load

P_cr = n·P = 2.5·45000 = 112500 N

KL = 2·2000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·112500·4000²/(π³·70000))^(1/4) = 85.35 mm

Answer

d = 85.35 mm

Check: the slenderness KL/r is 187.5, so the column is slender and Euler applies.

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