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 steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 500 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø40 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 500 mm

δ = mm

2.Axial stress & strainHard

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

P = 100 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø20 mm, 2 m, aluminium (E = 70 GPa)

δ = mm

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 1500 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

T = 1500 N·md_i/d_o = 0.6τ_allow = 75 MPa

d_o = mm

4.Torsion of shaftsHard

A hollow shaft transmits 30 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 30 kWN = 300 rpmd_o = 80 mmd_i = 60 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 2.5 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 = 65 kN at 2.5 mw = 3 kN/m from 0 to 2.5 m

M_A = 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.

25 kN @ 0 m50 kN @ 12 m60 kN @ 4 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 hollow rectangular (box) section has outer width B = 180 mm, outer height H = 275 mm and a uniform wall thickness t = 16 mm. Find its radius of gyration r about the horizontal centroidal axis.

B = 180 mmH = 275 mmt = 16 mm

r = mm

8.Centroid & second moment of areaHard

A welded I-section has a top flange 100 × 14 mm, a web 340 mm high and 8 mm thick, and a bottom flange 220 × 18 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 100 × 14 mmweb: 8 × 340 mmbottom flange: 220 × 18 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 50 MPa, σᵧ = −10 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 50 MPaσ_y = −10 MPaτ_xy = −30 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = 120 MPa and τₓᵧ = −50 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = 160 MPaσ_y = 120 MPaτ_xy = −50 MPa

σ_vM = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1.5 m from support A. Find the deflection δ_P under the load.

P = 3 kNa = 1.5 mL = 6 mE = 70 GPa (aluminium)I = 12 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 75 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/300. Find the smallest depth h_req of the section.

w = 1.5 kN/mL = 3.5 mE = 11 GPa (timber)b = 75 mm

h_req = mm

13.Column bucklingMedium

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

L = 6 mE = 70 GPa (aluminium)b × h = 50 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 75 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 75 kNn = 2.5L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.1668 mm

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·40²/4 = 1257 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 110·10³·5890 = 8.993 × 10⁸ N

Shortening

δ = 300000·500/8.993 × 10⁸ = 0.1668 mm

Answer

δ = 0.1668 mm

2.Axial stress & strain

Answer: δ = 11.92 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·30²/4 = 706.9 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 141.5 MPa

σ₂ = 318.3 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 141.5·2000/100000 = 2.829 mm

δ₂ = σ₂L₂/E₂ = 318.3·2000/70000 = 9.095 mm

δ = δ₁ + δ₂ = 11.92 mm

Answer

δ = 11.92 mm

3.Torsion of shafts

Answer: d_o = 48.91 mm

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.

Outer diameter from the allowable stress

T = 1500 N·m = 1.5 × 10⁶ N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·1.5 × 10⁶/(π·75·0.8704)) = 48.91 mm

d_i = 0.6·48.91 = 29.35 mm

Answer

d_o = 48.91 mm

A solid shaft would need d = 46.7 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: τ_max = 13.9 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 30000/31.42 = 954.9 N·m

Polar second moment of area

J = π(80⁴ − 60⁴)/32 = 2.749 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 954930·40/2.749 × 10⁶ = 13.9 MPa

Answer

τ_max = 13.9 MPa

5.Shear force & bending moment

Answer: M_A = −171.9 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: −65·2.5 = −162.5 kN·m

w: −3·2.5²/2 = −9.375 kN·m

M_A = −162.5 − 9.375 = −171.9 kN·m

Answer

M_A = −171.9 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: E (x = 8 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 = 40 kN, R₂ = 95 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 = −50, M_C = −20, M_D = −110, M_E = −200 kN·m, M_F = −100, M_G = 0

Answer

Mark E: M_E = −200 kN·m

7.Centroid & second moment of area

Answer: r = 99.86 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

b = 148 mm, h = 243 mm

A = BH − bh = 180·275 − 148·243 = 13536 mm²

I = (BH³ − bh³)/12 = 1.35 × 10⁸ mm⁴

Radius of gyration

r = √(I/A) = √(1.35 × 10⁸/13536) = 99.86 mm

Answer

r = 99.86 mm

8.Centroid & second moment of area

Answer: I_x = 170.8 × 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 three rectangles

Measure every y from the bottom edge.

top flange: A = 1400 mm², y = 365 mm

web: A = 2720 mm², y = 188 mm

bottom flange: A = 3960 mm², y = 9 mm

Centroid

ȳ = (1400·365 + 2720·188 + 3960·9)/8080 = 130.9 mm

Parallel-axis theorem

top flange: 100·14³/12 + 1400·234.1² = 7.672 × 10⁷ mm⁴

web: 8·340³/12 + 2720·57.06² = 3.506 × 10⁷ mm⁴

bottom flange: 220·18³/12 + 3960·121.9² = 5.899 × 10⁷ mm⁴

I_x = 7.672 × 10⁷ + 3.506 × 10⁷ + 5.899 × 10⁷

I_x = 1.708 × 10⁸ mm⁴ = 170.8 × 10⁶ mm⁴

Answer

I_x = 170.8 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = −67.5 °

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −30/42.43 = −0.7071, sin 2θ_s = −30/42.43 = −0.7071

2θ_s = atan2(−30, −30) = −135°

Angle on the element

θ_s = −67.5°

check: θ_s − θ_p = −67.5 − (−22.5) = −45°

Answer

θ_s = −67.5°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_vM = 168.2 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. 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 = 140 MPa, R = 53.85 MPa

σ₁ = 193.9 MPa, σ₂ = 86.15 MPa, σ₃ = 0

Von Mises

σ_vM = √(193.9² − 193.9·86.15 + 86.15²) = 168.2 MPa

Answer

σ_vM = 168.2 MPa

11.Beam deflection

Answer: δ_P = 9.04 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 = 3000 N, a = 1500 mm, b = 4500 mm, L = 6000 mm

EI = 8.4 × 10¹¹ N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 3000·1500²·4500²/(3·8.4 × 10¹¹·6000) = 9.04 mm

Answer

δ_P = 9.04 mm

12.Beam deflection

Answer: h_req = 154 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.

w = 1.5 N/mm, L = 3500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/300 = 3500/300 = 11.67 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·1.5·3500⁴/(384·11000·11.67) = 2.284 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·2.284 × 10⁷/75)^1/3 = 154 mm

Answer

h_req = 154 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 15.99 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 = 80·50³/12 = 833333 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·833333/6000² = 15992 N = 15.99 kN

Answer

P_cr = 15.99 kN

14.Column buckling

Answer: d = 74.59 mm

Required critical load

P_cr = n·P = 2.5·75000 = 187500 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·187500·4000²/(π³·200000))^(1/4) = 74.59 mm

Answer

d = 74.59 mm

Check: the slenderness KL/r is 214.5, so the column is slender and Euler applies.

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