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 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa). Both are 800 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 110 GPaL = 800 mm

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø120/Ø50 mm, E = 100 GPa, σ_allow = 100 MPa

P_max = kN

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 60 MPa. Find the smallest outer diameter d_o.

T = 1500 N·md_i/d_o = 0.6τ_allow = 60 MPa

d_o = mm

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 5 kWN = 500 rpmτ_allow = 60 MPa

d = mm

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 @ 8 m4 kN/m, 0–12 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

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 @ 12 m60 kN @ 4 m10 kN/m, 0–10 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 100 mm × 200 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 80 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 200 mmd = 30 mmhole centre: y = 80 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A welded I-section has a top flange 130 × 10 mm, a web 270 mm high and 12 mm thick, and a bottom flange 300 × 22 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 130 × 10 mmweb: 12 × 270 mmbottom flange: 300 × 22 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 30 MPa and σᵧ = 70 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 125 MPa?

σ_x = 30 MPaσ_y = 70 MPaσ₁ = 125 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 20 MPa, σᵧ = −60 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σ_x = 20 MPaσ_y = −60 MPaτ_xy = −30 MPaθ = 50°

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 = 2 kN/mL = 4 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_max = 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 = 12 kN/mL = 1.5 mE = 200 GPa (steel)I = 8.5 × 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 = 6 mE = 200 GPa (steel)b × h = 40 × 50 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 125 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 125 kNn = 3L = 1 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.06143 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·25²/4 = 490.9 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·5027 = 6.511 × 10⁸ N

Shortening

δ = 50000·800/6.511 × 10⁸ = 0.06143 mm

Answer

δ = 0.06143 mm

2.Axial stress & strain

Answer: P_max = 774.6 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·9346 = 1.033 × 10⁹ N

Governing strain

ε = min(0.00075, 0.001) = 0.00075 (steel)

Largest load

P_max = 0.00075·1.033 × 10⁹ = 774599 N = 774.6 kN

Answer

P_max = 774.6 kN

3.Torsion of shafts

Answer: d_o = 52.69 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·60·0.8704)) = 52.69 mm

d_i = 0.6·52.69 = 31.61 mm

Answer

d_o = 52.69 mm

A solid shaft would need d = 50.31 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 20.09 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 5000/52.36 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·60)) = 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: 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 = 39 kN, R₂ = 54 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 = 70, M_C = 124, M_D = 162, M_E = 184 kN·m, M_F = 100, M_G = 0

Answer

Mark E: M_E = 184 kN·m

6.Shear force & bending moment

Answer: C (x = 4 m)

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 75 kN, R₂ = 140 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 = 130, M_C = 220 kN·m, M_D = 150, M_E = 40, M_F = −110, M_G = 0

Answer

Mark C: M_C = 220 kN·m

7.Centroid & second moment of area

Answer: I_x = 66.33 × 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 = (20000·100 − 706.9·80)/(20000 − 706.9) = 100.7 mm

Full plate about ȳ

100·200³/12 + 20000·−0.7328² = 66.68 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−20.73² = 0.3436 × 10⁶ mm⁴

Subtract

I_x = 66.33 × 10⁶ mm⁴

Answer

I_x = 66.33 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 131.3 × 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 = 1300 mm², y = 297 mm

web: A = 3240 mm², y = 157 mm

bottom flange: A = 6600 mm², y = 11 mm

Centroid

ȳ = (1300·297 + 3240·157 + 6600·11)/11140 = 86.84 mm

Parallel-axis theorem

top flange: 130·10³/12 + 1300·210.2² = 5.743 × 10⁷ mm⁴

web: 12·270³/12 + 3240·70.16² = 3.563 × 10⁷ mm⁴

bottom flange: 300·22³/12 + 6600·75.84² = 3.823 × 10⁷ mm⁴

I_x = 5.743 × 10⁷ + 3.563 × 10⁷ + 3.823 × 10⁷

I_x = 1.313 × 10⁸ mm⁴ = 131.3 × 10⁶ mm⁴

Answer

I_x = 131.3 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_xy = 72.28 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 = (30 + 70)/2 = 50 MPa

R = σ₁ − C = 125 − 50 = 75 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(75² − 20²) = 72.28 MPa

Answer

τ_xy = 72.28 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −56.49 MPa

Transformation equation

σ_θ = −20 + 40·−0.1736 + −30·0.9848 = −56.49 MPa

Answer

σ_θ = −56.49 MPa

11.Beam deflection

Answer: δ_max = 1.671 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 = 2 N/mm, L = 4000 mm

EI = 3.99 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·2·4000⁴/(384·3.99 × 10¹²) = 1.671 mm

Answer

δ_max = 1.671 mm

12.Beam deflection

Answer: δ_max = 1.417 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 = 12 N/mm, L = 1500 mm, EI = 1.7 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 23000·1500³/(48·1.7 × 10¹²) = 0.9513 mm

δ_w = 5wL⁴/(384EI) = 5·12·1500⁴/(384·1.7 × 10¹²) = 0.4653 mm

Add them

δ_max = 0.9513 + 0.4653 = 1.417 mm

Answer

δ_max = 1.417 mm

13.Column buckling

Answer: P_cr = 3.655 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 = 2·6000 = 12000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·266667/12000² = 3655 N = 3.655 kN

Answer

P_cr = 3.655 kN

14.Column buckling

Answer: d = 44.35 mm

Required critical load

P_cr = n·P = 3·125000 = 375000 N

KL = 1·1000 = 1000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·375000·1000²/(π³·200000))^(1/4) = 44.35 mm

Answer

d = 44.35 mm

Check: the slenderness KL/r is 90.18, so the column is slender and Euler applies.

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