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

Find the elongation δ of the brass bar under the axial load.

P = 35 kNd = 12 mmL = 1.5 mE = 100 GPa

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 = 105 kNsegment 1: Ø25 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid steel shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 100 N·md = 50 mmL = 3 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1500 rpmd_o = 100 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 3 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 = 3 mP = 55 kN at 3 mw = 2 kN/m from 0 to 3 m

M_A = kN·m

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment M_max in the beam.

L = 10 mP = 70 kN at 8 mw = 5 kN/m from 0 to 10 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 200 mm × 240 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 72 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 240 mmd = 50 mmhole centre: y = 72 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 15 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 100 mmt_f = 10 mmt_w = 12 mmh_w = 280 mmM = 15 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 30 MPa, σᵧ = 10 MPa and τₓᵧ = −60 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 = 30 MPaσ_y = 10 MPaτ_xy = −60 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 600 mm and wall thickness 6 mm holds gas at a gauge pressure of 2 MPa. It is made from plate welded along a helical seam at 30° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 600 mmt = 6 mmp = 2 MPaα = 30°

σ_n = 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 = 5 kN/mL = 2.5 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the reaction R_B at the prop.

L = 4 mP = 50 kN at 2 m

R_B = kN

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 = 1.5 mE = 70 GPa (aluminium)b × h = 30 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 55 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 55 kNn = 3L = 5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 4.642 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·12²/4 = 113.1 mm²

σ = P/A = 35·1000/113.1 = 309.5 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 309.5/100000 = 0.003095 = 3095 µε

Elongation

δ = ε·L = 0.003095·1500 mm = 4.642 mm

Answer

δ = 4.642 mm

2.Axial stress & strain

Answer: σ_max = 928.4 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·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 213.9 MPa

σ₂ = 928.4 MPa

Largest stress

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

σ_max = 928.4 MPa

Answer

σ_max = 928.4 MPa

3.Torsion of shafts

Answer: φ = 0.3502 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·3000/(80000·613592) = 0.006112 rad

φ = 0.006112·180/π = 0.3502°

Answer

φ = 0.3502°

4.Torsion of shafts

Answer: τ_max = 3.725 MPa

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 100000/157.1 = 636.6 N·m

Polar second moment of area

J = π(100⁴ − 60⁴)/32 = 8.545 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 636620·50/8.545 × 10⁶ = 3.725 MPa

Answer

τ_max = 3.725 MPa

5.Shear force & bending moment

Answer: M_A = −174 kN·m

Keep the free-end part

P: −55·3 = −165 kN·m

w: −2·3²/2 = −9 kN·m

M_A = −165 − 9 = −174 kN·m

Answer

M_A = −174 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: M_max = 152.1 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·10 = 50 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 70·8 + 5·10·5

R_B = 810/10 = 81 kN

ΣF_y = 0: R_A = 120 − 81 = 39 kN

Where is V = 0?

R_A − w·x₀ = 0 → x₀ = 39/5 = 7.8 m

Maximum moment

M_max = R_A·x₀ − w·x₀²/2 = 39·7.8 − 5·7.8²/2 = 152.1 kN·m

Answer

M_max = 152.1 kN·m

It occurs at x = 7.8 m from A.

7.Centroid & second moment of area

Answer: I_x = 225.4 × 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 = (48000·120 − 1963·72)/(48000 − 1963) = 122 mm

Full plate about ȳ

200·240³/12 + 48000·−2.047² = 230.6 × 10⁶ mm⁴

Hole about ȳ

π·50⁴/64 + 1963·−50.05² = 5.225 × 10⁶ mm⁴

Subtract

I_x = 225.4 × 10⁶ mm⁴

Answer

I_x = 225.4 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 68.1 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 = 100·10 = 1000 mm², y_f = 280 + 10/2 = 285 mm

web: A_w = 12·280 = 3360 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1000·285 + 3360·140)/4360 = 173.3 mm

Parallel-axis theorem

flange: 100·10³/12 + 1000·(285 − 173.3)² = 1.249 × 10⁷ mm⁴

web: 12·280³/12 + 3360·(140 − 173.3)² = 2.567 × 10⁷ mm⁴

I_x = 1.249 × 10⁷ + 2.567 × 10⁷ = 3.816 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 173.3 mm

σ = M·y/I = 1.5 × 10⁷·173.3/3.816 × 10⁷ = 68.1 MPa

Answer

σ_bottom = 68.1 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −40.27 °

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·−60, 30 − 10) = −80.54°

Angle on the element

θ_p = −40.27°

Answer

θ_p = −40.27°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_n = 87.5 MPa

Wall stresses

r = 600/2 = 300 mm

σ_h = pr/t = 2·300/6 = 100 MPa

σ_a = pr/(2t) = 50 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.

σ_n = 50·sin²30° + 100·cos²30° = 87.5 MPa

Answer

σ_n = 87.5 MPa

11.Beam deflection

Answer: δ_max = 4.056 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 = 5 N/mm, L = 2500 mm

EI = 6.27 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·5·2500⁴/(384·6.27 × 10¹¹) = 4.056 mm

Answer

δ_max = 4.056 mm

12.Beam deflection

Answer: R_B = 15.63 kN

Remove the prop

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

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

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) = 5PL³/(48EI) → R_B = 5P/16 = 5·50/16 = 15.63 kN

Answer

R_B = 15.63 kN

13.Column buckling

Answer: P_cr = 41.45 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·135000/1500² = 41452 N = 41.45 kN

Answer

P_cr = 41.45 kN

14.Column buckling

Answer: d = 114.2 mm

Required critical load

P_cr = n·P = 3·55000 = 165000 N

KL = 2·5000 = 10000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·165000·10000²/(π³·200000))^(1/4) = 114.2 mm

Answer

d = 114.2 mm

Check: the slenderness KL/r is 350.2, so the column is slender and Euler applies.

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