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 aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa). Both are 600 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø40 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 70 GPaL = 600 mm

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 115 kNsegment 1: Ø30 mm, 1 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 60 mm, inner diameter 45 mm, length 1.5 m) is fixed at one end and twisted by a torque of 2000 N·m at the other. Find the angle of twist φ in degrees.

T = 2000 N·md_o = 60 mmd_i = 45 mmL = 1.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 15 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 15 kWN = 300 rpmd_o = 80 mmd_i = 40 mm

τ_max = MPa

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.

10 kN @ 8 m45 kN @ 2 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)

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 10 kN at 6 mw = 5 kN/m from 0 to 8 mx = 2.5 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 = 10 mmt_w = 16 mmh_w = 160 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 130 × 24 mm, a web 310 mm high and 14 mm thick, and a bottom flange 240 × 18 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 130 × 24 mmweb: 14 × 310 mmbottom flange: 240 × 18 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 130 MPa, σᵧ = 0 MPa and τₓᵧ = −60 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 = 130 MPaσ_y = 0 MPaτ_xy = −60 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 160 MPa, σᵧ = −30 MPa and τₓᵧ = 50 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 30° clockwise from the x-axis.

σ_x = 160 MPaσ_y = −30 MPaτ_xy = 50 MPaθ = -30°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.

P = 24 kNw = 5 kN/mL = 2 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 8 kN/mL = 1.5 mE = 11 GPa (timber)

I_req = × 10⁶ mm⁴

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 = 1.5 mE = 200 GPa (steel)b × h = 40 × 50 mm

P_cr = kN

14.Column bucklingHard

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

L = 3 mE = 70 GPa (aluminium)d = 20 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.03878 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(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 70·10³·7461 = 7.736 × 10⁸ N

Shortening

δ = 50000·600/7.736 × 10⁸ = 0.03878 mm

Answer

δ = 0.03878 mm

2.Axial stress & strain

Answer: σ_max = 1017 MPa

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·12²/4 = 113.1 mm²

σ₁ = 162.7 MPa

σ₂ = 1017 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 1017 MPa

Answer

σ_max = 1017 MPa

3.Torsion of shafts

Answer: φ = 5.067 °

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(60⁴ − 45⁴)/32 = 869767 mm⁴

Angle of twist

T = 2000 N·m = 2 × 10⁶ N·mm

L = 1.5 m = 1500 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 2 × 10⁶·1500/(39000·869767) = 0.08844 rad

φ = 0.08844·180/π = 5.067°

Answer

φ = 5.067°

4.Torsion of shafts

Answer: τ_max = 5.066 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 15000/31.42 = 477.5 N·m

Polar second moment of area

J = π(80⁴ − 40⁴)/32 = 3.77 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 477465·40/3.77 × 10⁶ = 5.066 MPa

Answer

τ_max = 5.066 MPa

5.Shear force & bending moment

Answer: B (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 = 40.83 kN, R₂ = 14.17 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 = 81.67 kN·m, M_C = 73.33, M_D = 65, M_E = 56.67, M_F = 28.33, M_G = 0

Answer

Mark B: M_B = 81.67 kN·m

6.Shear force & bending moment

Answer: M_x = 40.63 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 = 5·8 = 40 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 10·6 + 5·8·4

R_B = 220/8 = 27.5 kN

ΣF_y = 0: R_A = 50 − 27.5 = 22.5 kN

Cut at x = 2.5 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 = 22.5·2.5 − 5·2.5·1.25 = 40.63 kN·m

Answer

M_x = 40.63 kN·m

7.Centroid & second moment of area

Answer: I_x = 12.01 × 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 = 140·10 = 1400 mm², y_f = 160 + 10/2 = 165 mm

web: A_w = 16·160 = 2560 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1400·165 + 2560·80)/3960 = 110.1 mm

Parallel-axis theorem

flange: 140·10³/12 + 1400·(165 − 110.1)² = 4.239 × 10⁶ mm⁴

web: 16·160³/12 + 2560·(80 − 110.1)² = 7.773 × 10⁶ mm⁴

I_x = 4.239 × 10⁶ + 7.773 × 10⁶ = 1.201 × 10⁷ mm⁴ = 12.01 × 10⁶ mm⁴

Answer

I_x = 12.01 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 235.2 × 10⁶ mm⁴

Split into three rectangles

Measure every y from the bottom edge.

top flange: A = 3120 mm², y = 340 mm

web: A = 4340 mm², y = 173 mm

bottom flange: A = 4320 mm², y = 9 mm

Centroid

ȳ = (3120·340 + 4340·173 + 4320·9)/11780 = 157.1 mm

Parallel-axis theorem

top flange: 130·24³/12 + 3120·182.9² = 1.045 × 10⁸ mm⁴

web: 14·310³/12 + 4340·15.91² = 3.585 × 10⁷ mm⁴

bottom flange: 240·18³/12 + 4320·148.1² = 9.485 × 10⁷ mm⁴

I_x = 1.045 × 10⁸ + 3.585 × 10⁷ + 9.485 × 10⁷

I_x = 2.352 × 10⁸ mm⁴ = 235.2 × 10⁶ mm⁴

Answer

I_x = 235.2 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_s = −66.35 °

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.

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 = −60/88.46 = −0.6783, sin 2θ_s = −65/88.46 = −0.7348

2θ_s = atan2(−65, −60) = −132.7°

Angle on the element

θ_s = −66.35°

check: θ_s − θ_p = −66.35 − (−21.35) = −45°

Answer

θ_s = −66.35°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 69.2 MPa

Transformation equation

σ_θ = 65 + 95·0.5 + 50·−0.866 = 69.2 MPa

Answer

σ_θ = 69.2 MPa

11.Beam deflection

Answer: δ = 30.83 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 = 24000 N, w = 5 N/mm, L = 2000 mm

EI = 2.4 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 24000·2000³/(3·2.4 × 10¹²) = 26.67 mm

δ_w = wL⁴/(8EI) = 5·2000⁴/(8·2.4 × 10¹²) = 4.167 mm

Add them

δ = 26.67 + 4.167 = 30.83 mm

Answer

δ = 30.83 mm

12.Beam deflection

Answer: I_req = 11.51 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 1500/360 = 4.167 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·8·1500⁴/(384·11000·4.167) = 1.151 × 10⁷ mm⁴

I_req = 11.51 × 10⁶ mm⁴

Answer

I_req = 11.51 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 58.49 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 = 50·40³/12 = 266667 mm⁴

Effective length

KL = 2·1500 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·266667/3000² = 58487 N = 58.49 kN

Answer

P_cr = 58.49 kN

14.Column buckling

Answer: σ_cr = 7.676 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/5 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa

Answer

σ_cr = 7.676 MPa

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