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 bar is made of a steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: copper, Ø60 mm, 500 mm, E = 120 GPaδ = 0.3 mm

P = kN

2.Axial stress & strainHard

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

P = 120 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, copper (E = 117 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel shaft 1.2 m long carries a torque of 250 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.

T = 250 N·mL = 1.2 mG = 80 GPaφ_allow = 1°

d = mm

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 500 rpmd_o = 50 mmd_i = 35 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.

45 kN @ 3 m6 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 = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mP = 50 kN at 3 mw = 8 kN/m from 0 to 4 mx = 2 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 120 mm × 200 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 60 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 200 mmd = 40 mmhole centre: y = 60 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

The T-section is used as a beam with the flange on top and carries a sagging bending moment of 25 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 100 mmt_f = 25 mmt_w = 18 mmh_w = 120 mmM = 25 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 120 MPa, σᵧ = −20 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 120 MPaσ_y = −20 MPaτ_xy = −50 MPa

θ_p = °

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 3 MPa. It is made from plate welded along a helical seam at 60° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 400 mmt = 5 mmp = 3 MPaα = 60°

|τ_n| = 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 = 21 kNw = 10 kN/mL = 1.5 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ = 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 = 16 kNw = 15 kN/mL = 3 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 4 mE = 200 GPa (steel)b × h = 50 × 60 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 = 5 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 27.22 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

A₁ = 314.2 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·314.2) = 9.549 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(120000·2827) = 1.474 × 10⁻⁶ mm/N

Load

P = 0.3/1.102 × 10⁻⁵ = 27216 N = 27.22 kN

Answer

P = 27.22 kN

2.Axial stress & strain

Answer: δ = 5.216 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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 95.49 MPa

σ₂ = 1061 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 95.49·500/70000 = 0.6821 mm

δ₂ = σ₂L₂/E₂ = 1061·500/117000 = 4.534 mm

δ = δ₁ + δ₂ = 5.216 mm

Answer

δ = 5.216 mm

3.Torsion of shafts

Answer: d = 38.46 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.

Required polar moment

T = 250 N·m = 250000 N·mm

L = 1.2 m = 1200 mm

G = 80 GPa = 80000 MPa

φ = 1·π/180 = 0.01745 rad

J = TL/(Gφ) = 250000·1200/(80000·0.01745) = 214859 mm⁴

Diameter

d = (32J/π)^1/4 = 38.46 mm

Answer

d = 38.46 mm

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

4.Torsion of shafts

Answer: τ_max = 20.48 MPa

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 20000/52.36 = 382 N·m

Polar second moment of area

J = π(50⁴ − 35⁴)/32 = 466269 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 381972·25/466269 = 20.48 MPa

Answer

τ_max = 20.48 MPa

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 = 40.5 kN, R₂ = 40.5 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 = 37.5, M_C = 69, M_D = 94.5 kN·m, M_E = 69, M_F = 37.5, M_G = 0

Answer

Mark D: M_D = 94.5 kN·m

6.Shear force & bending moment

Answer: M_x = 41 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 = 8·4 = 32 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 50·3 + 8·4·2

R_B = 214/4 = 53.5 kN

ΣF_y = 0: R_A = 82 − 53.5 = 28.5 kN

Cut at x = 2 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 = 28.5·2 − 8·2·1 = 41 kN·m

Answer

M_x = 41 kN·m

7.Centroid & second moment of area

Answer: I_x = 77.75 × 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 = (24000·100 − 1257·60)/(24000 − 1257) = 102.2 mm

Full plate about ȳ

120·200³/12 + 24000·−2.21² = 80.12 × 10⁶ mm⁴

Hole about ȳ

π·40⁴/64 + 1257·−42.21² = 2.365 × 10⁶ mm⁴

Subtract

I_x = 77.75 × 10⁶ mm⁴

Answer

I_x = 77.75 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 130.8 MPa

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 = 100·25 = 2500 mm², y_f = 120 + 25/2 = 132.5 mm

web: A_w = 18·120 = 2160 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2500·132.5 + 2160·60)/4660 = 98.89 mm

Parallel-axis theorem

flange: 100·25³/12 + 2500·(132.5 − 98.89)² = 2.953 × 10⁶ mm⁴

web: 18·120³/12 + 2160·(60 − 98.89)² = 5.86 × 10⁶ mm⁴

I_x = 2.953 × 10⁶ + 5.86 × 10⁶ = 8.813 × 10⁶ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 145 − 98.89 = 46.11 mm

σ = M·y/I = 2.5 × 10⁷·46.11/8.813 × 10⁶ = 130.8 MPa

Answer

σ_top = 130.8 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −17.77 °

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

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·−50, 120 − −20) = −35.54°

Angle on the element

θ_p = −17.77°

Answer

θ_p = −17.77°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 25.98 MPa

Wall stresses

r = 400/2 = 200 mm

σ_h = pr/t = 3·200/5 = 120 MPa

σ_a = pr/(2t) = 60 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.

|τ_n| = (120 − 60)·sin 60°·cos 60° = 25.98 MPa

Answer

|τ_n| = 25.98 MPa

11.Beam deflection

Answer: δ = 7.507 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 = 21000 N, w = 10 N/mm, L = 1500 mm

EI = 3.99 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 21000·1500³/(3·3.99 × 10¹²) = 5.921 mm

δ_w = wL⁴/(8EI) = 10·1500⁴/(8·3.99 × 10¹²) = 1.586 mm

Add them

δ = 5.921 + 1.586 = 7.507 mm

Answer

δ = 7.507 mm

12.Beam deflection

Answer: δ_max = 18.8 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 = 16000 N, w = 15 N/mm, L = 3000 mm, EI = 1.32 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 16000·3000³/(48·1.32 × 10¹²) = 6.818 mm

δ_w = 5wL⁴/(384EI) = 5·15·3000⁴/(384·1.32 × 10¹²) = 11.99 mm

Add them

δ_max = 6.818 + 11.99 = 18.8 mm

Answer

δ_max = 18.8 mm

13.Column buckling

Answer: P_cr = 77.11 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 = 60·50³/12 = 625000 mm⁴

Effective length

KL = 1·4000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·625000/4000² = 77106 N = 77.11 kN

Answer

P_cr = 77.11 kN

14.Column buckling

Answer: σ_cr = 4.441 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 1·5000 = 5000 mm

KL/r = 5000/7.5 = 666.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/666.7² = 4.441 MPa

Answer

σ_cr = 4.441 MPa

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