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 30 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 50 mm, length 300 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: copper, Ø50 mm, 300 mm, E = 120 GPaδ = 0.2 mm

P = kN

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 = 35 kNsegment 1: Ø40 mm, 2 m, copper (E = 117 GPa)segment 2: Ø12 mm, 2 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 400 N·m. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter d_o.

T = 400 N·md_i/d_o = 0.5τ_allow = 80 MPa

d_o = mm

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 1500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1500 rpmτ_allow = 50 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 30 kN·m at x = 3 m from A, and no other load. Find the bending moment M_x just to the left of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 4 mM₀ = 30 kN·m clockwise, at 3 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 7 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 55 kN at 7 mw = 12 kN/m from 0 to 8 mx = 7 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 100 mm × 150 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 52.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 150 mmd = 40 mmhole centre: y = 52.5 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 200 mm plate has a rectangular cut-out 60 mm wide and 20 mm high, centred on the 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 = 200 mmh = 200 mmcut-out: 60 × 20 mmhole centre: y = 60 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 0 MPa, σᵧ = −10 MPa and τₓᵧ = 20 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 0 MPaσ_y = −10 MPaτ_xy = 20 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −50 MPa, σᵧ = 120 MPa and τₓᵧ = −50 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −50 MPaσ_y = 120 MPaτ_xy = −50 MPaσ_Y = 350 MPa

n =

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 = 20 kN/mL = 2 mE = 70 GPa (aluminium)I = 23 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 8 mw = 6 kN/m

R_B = kN

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

14.Column bucklingHard

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

L = 6 mE = 70 GPa (aluminium)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 48.75 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 = 706.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·706.9) = 2.829 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(120000·1963) = 1.273 × 10⁻⁶ mm/N

Load

P = 0.2/4.103 × 10⁻⁶ = 48749 N = 48.75 kN

Answer

P = 48.75 kN

2.Axial stress & strain

Answer: σ_max = 309.5 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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 27.85 MPa

σ₂ = 309.5 MPa

Largest stress

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

σ_max = 309.5 MPa

Answer

σ_max = 309.5 MPa

3.Torsion of shafts

Answer: d_o = 30.06 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 = 400 N·m = 400000 N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·400000/(π·80·0.9375)) = 30.06 mm

d_i = 0.5·30.06 = 15.03 mm

Answer

d_o = 30.06 mm

A solid shaft would need d = 29.42 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 21.35 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 15000/157.1 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·50)) = 21.35 mm

Answer

d = 21.35 mm

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

5.Shear force & bending moment

Answer: M_x = −22.5 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 30/4 = 7.5 kN (upward)

R_A = −R_B = −7.5 kN (downward)

Cut just left of the couple

The left part contains only R_A.

M_x = R_A·a = −7.5·3 = −22.5 kN·m

Answer

M_x = −22.5 kN·m

6.Shear force & bending moment

Answer: M_x = 90.13 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·8 = 96 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 55·7 + 12·8·4

R_B = 769/8 = 96.13 kN

ΣF_y = 0: R_A = 151 − 96.13 = 54.88 kN

Cut at x = 7 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 = 54.88·7 − 12·7·3.5 = 90.13 kN·m

Answer

M_x = 90.13 kN·m

7.Centroid & second moment of area

Answer: I_x = 27.3 × 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 = (15000·75 − 1257·52.5)/(15000 − 1257) = 77.06 mm

Full plate about ȳ

100·150³/12 + 15000·−2.057² = 28.19 × 10⁶ mm⁴

Hole about ȳ

π·40⁴/64 + 1257·−24.56² = 0.8835 × 10⁶ mm⁴

Subtract

I_x = 27.3 × 10⁶ mm⁴

Answer

I_x = 27.3 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 131.3 × 10⁶ mm⁴

Centroid

ȳ = (40000·100 − 1200·60)/(40000 − 1200) = 101.2 mm

Full plate about ȳ

200·200³/12 + 40000·−1.237² = 133.4 × 10⁶ mm⁴

Hole about ȳ

60·20³/12 + 1200·−41.24² = 2.081 × 10⁶ mm⁴

Subtract

I_x = 131.3 × 10⁶ mm⁴

Answer

I_x = 131.3 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_abs = 20.62 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = −5 MPa, R = √(5² + 20²) = 20.62 MPa

σ₁ = 15.62 MPa, σ₂ = −25.62 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(41.23, 15.62, 25.62)/2 = 20.62 MPa

Answer

τ_abs = 20.62 MPa

Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.

10.Mohr’s circle & stress transformation

Answer: n = 2.007

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 = 35 MPa, R = 98.62 MPa

σ₁ = 133.6 MPa, σ₂ = −63.62 MPa, σ₃ = 0

Von Mises

σ_vM = √(133.6² − 133.6·(−63.62) + (−63.62)²) = 174.4 MPa

n = 350/174.4 = 2.007

Answer

n = 2.007

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

11.Beam deflection

Answer: δ_max = 2.588 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 = 20 N/mm, L = 2000 mm

EI = 1.61 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·20·2000⁴/(384·1.61 × 10¹²) = 2.588 mm

Answer

δ_max = 2.588 mm

12.Beam deflection

Answer: R_B = 18 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·6·8/8 = 18 kN

Answer

R_B = 18 kN

13.Column buckling

Answer: P_cr = 177.2 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 = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2800² = 177179 N = 177.2 kN

Answer

P_cr = 177.2 kN

14.Column buckling

Answer: σ_cr = 1.919 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 2·6000 = 12000 mm

KL/r = 12000/20 = 600

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/600² = 1.919 MPa

Answer

σ_cr = 1.919 MPa

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