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 = 95 kNd = 10 mmL = 3 mE = 117 GPa

ε = µε

2.Axial stress & strainHard

A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of aluminium (diameter 50 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 500 mm, E = 70 GPaP = 20 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 = 100 N·md = 40 mmL = 2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 1800 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 1800 rpmd_o = 60 mmd_i = 42 mm

τ_max = MPa

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 6.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 5 kN/m from 0 to 10 mx = 6.5 m

M_x = kN·m

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.

55 kN @ 6 m60 kN @ 2 m5 kN/m, 0–5 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)

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 = 220 mmt_f = 20 mmt_w = 14 mmh_w = 160 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 90 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 240 mmt_f = 20 mmt_w = 10 mmh_w = 120 mmσ_t,allow = 40 MPaσ_c,allow = 90 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 60 MPa and σᵧ = 80 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 115 MPa?

σ_x = 60 MPaσ_y = 80 MPaσ₁ = 115 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 400 mm and wall thickness 5 mm holds gas at a gauge pressure of 1 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 400 mmt = 5 mmp = 1 MPaα = 55°

σ_n = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_B at the free end B.

P = 5 kNa = 1 mL = 3 mE = 11 GPa (timber)I = 8.5 × 10⁶ mm⁴

δ_B = 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 = 23 kNw = 20 kN/mL = 4 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 6 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium 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 = 4 mE = 70 GPa (aluminium)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 10338 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 = 95·1000/78.54 = 1210 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 1210/117000 = 0.01034 = 10338 µε

Answer

ε = 10338 µε

2.Axial stress & strain

Answer: σ₂ = 5.942 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 = 200000·490.9/500 = 196350 N/mm

k₂ = 70000·1963/500 = 274889 N/mm

Share of segment 2

F₂ = 20000·274889/471239 = 11667 N

Stress

σ₂ = 11667/1963 = 5.942 MPa (compression)

Answer

σ₂ = 5.942 MPa

3.Torsion of shafts

Answer: φ = 1.754 °

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) = 100·1000·2000/(26000·251327) = 0.03061 rad

φ = 0.03061·180/π = 1.754°

Answer

φ = 1.754°

4.Torsion of shafts

Answer: τ_max = 3.292 MPa

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 20000/188.5 = 106.1 N·m

Polar second moment of area

J = π(60⁴ − 42⁴)/32 = 966855 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 106103·30/966855 = 3.292 MPa

Answer

τ_max = 3.292 MPa

5.Shear force & bending moment

Answer: M_x = 56.88 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 = 5·10 = 50 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 5·10·5

R_B = 250/10 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 6.5 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 = 25·6.5 − 5·6.5·3.25 = 56.88 kN·m

Answer

M_x = 56.88 kN·m

6.Shear force & bending moment

Answer: C (x = 2 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 = 37.5 kN, R₂ = 102.5 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 = 35, M_C = 65 kN·m, M_D = 30, M_E = −10, M_F = −55, M_G = 0

Answer

Mark C: M_C = 65 kN·m

7.Centroid & second moment of area

Answer: I_x = 16.95 × 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 = 220·20 = 4400 mm², y_f = 160 + 20/2 = 170 mm

web: A_w = 14·160 = 2240 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4400·170 + 2240·80)/6640 = 139.6 mm

Parallel-axis theorem

flange: 220·20³/12 + 4400·(170 − 139.6)² = 4.203 × 10⁶ mm⁴

web: 14·160³/12 + 2240·(80 − 139.6)² = 1.275 × 10⁷ mm⁴

I_x = 4.203 × 10⁶ + 1.275 × 10⁷ = 1.695 × 10⁷ mm⁴ = 16.95 × 10⁶ mm⁴

Answer

I_x = 16.95 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 2.174 kN·m

Centroid

ȳ = (4800·130 + 1200·60)/6000 = 116 mm

top edge: c = 140 − 116 = 24 mm

Second moment of area

I = 240·20³/12 + 4800·14² + 10·120³/12 + 1200·56² = 6.304 × 10⁶ mm⁴

Moment allowed by each edge

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

top edge: M = 90·6.304 × 10⁶/24 = 23.64 kN·m

bottom edge: M = 40·6.304 × 10⁶/116 = 2.174 kN·m

Governing value

M_allow = min = 2.174 kN·m (bottom edge)

Answer

M_allow = 2.174 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_xy = 43.87 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.

Required radius

C = (60 + 80)/2 = 70 MPa

R = σ₁ − C = 115 − 70 = 45 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(45² − 10²) = 43.87 MPa

Answer

τ_xy = 43.87 MPa

10.Mohr’s circle & stress transformation

Answer: σ_n = 26.58 MPa

Wall stresses

r = 400/2 = 200 mm

σ_h = pr/t = 1·200/5 = 40 MPa

σ_a = pr/(2t) = 20 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 20·sin²55° + 40·cos²55° = 26.58 MPa

Answer

σ_n = 26.58 MPa

11.Beam deflection

Answer: δ_B = 71.3 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 = 5000 N, a = 1000 mm, L = 3000 mm

EI = 9.35 × 10¹⁰ N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 17.83 mm

θ_P = Pa²/(2EI) = 0.02674 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 71.3 mm

Answer

δ_B = 71.3 mm

12.Beam deflection

Answer: δ_max = 38.62 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 = 23000 N, w = 20 N/mm, L = 4000 mm, EI = 2.52 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 23000·4000³/(48·2.52 × 10¹²) = 12.17 mm

δ_w = 5wL⁴/(384EI) = 5·20·4000⁴/(384·2.52 × 10¹²) = 26.46 mm

Add them

δ_max = 12.17 + 26.46 = 38.62 mm

Answer

δ_max = 38.62 mm

13.Column buckling

Answer: P_cr = 441 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 = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/3000² = 440978 N = 441 kN

Answer

P_cr = 441 kN

14.Column buckling

Answer: σ_cr = 13.77 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 0.7·4000 = 2800 mm

KL/r = 2800/12.5 = 224

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/224² = 13.77 MPa

Answer

σ_cr = 13.77 MPa

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