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 axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).

P = 50 kNd = 25 mmL = 2.5 mE = 200 GPa

ε = µε

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 = 55 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1.5 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid aluminium 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 = 400 N·md = 30 mmL = 1 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 1800 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 5 kWN = 1800 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. It carries a point load P₁ between the supports and a point load P₂ at C. Find the shear force V_x just to the left of support B. Use sagging moments positive and upward shear on the left face positive.

L = 7 mP₁ = 35 kN at 2 mP₂ = 25 kN at 7 msupports at x = 0 and x = 5 m

V_x = kN

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 4 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 40 kN at 4 mw = 10 kN/m from 0 to 10 mx = 4 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 240 mm and inside diameter d = 208 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 240 mmd = 208 mm

r = mm

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 40 MPa in tension and 90 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 120 mmt_f = 10 mmt_w = 8 mmh_w = 200 mmσ_t,allow = 40 MPaσ_c,allow = 90 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 40 MPa, σᵧ = 110 MPa and τₓᵧ = 10 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 40 MPaσ_y = 110 MPaτ_xy = 10 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1200 mm and wall thickness 5 mm holds gas at a gauge pressure of 2 MPa. It is made from plate welded along a helical seam at 25° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1200 mmt = 5 mmp = 2 MPaα = 25°

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 = 20 kN/mL = 4 mE = 200 GPa (steel)I = 23 × 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 = 24 kNw = 4 kN/mL = 3 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_max = mm

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 = 2 mE = 70 GPa (aluminium)b × h = 20 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 85 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 85 kNn = 3L = 6 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 509.3 µε

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·25²/4 = 490.9 mm²

σ = P/A = 50·1000/490.9 = 101.9 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 101.9/200000 = 0.0005093 = 509.3 µε

Answer

ε = 509.3 µε

2.Axial stress & strain

Answer: σ_max = 486.3 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 = 77.81 MPa

σ₂ = 486.3 MPa

Largest stress

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

σ_max = 486.3 MPa

Answer

σ_max = 486.3 MPa

3.Torsion of shafts

Answer: φ = 11.08 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·1000/(26000·79522) = 0.1935 rad

φ = 0.1935·180/π = 11.08°

Answer

φ = 11.08°

4.Torsion of shafts

Answer: d = 13.11 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 5000/188.5 = 26.53 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·26526/(π·60)) = 13.11 mm

Answer

d = 13.11 mm

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

5.Shear force & bending moment

Answer: V_x = −24 kN

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. The overhang load also acts about A.

ΣM_A = 0: R_B·5 = 35·2 + 25·7

R_B = 245/5 = 49 kN

ΣF_y = 0: R_A = 60 − 49 = 11 kN

Cut just left of B and keep the left part

V_x = R_A − P₁ = 11 − 35 = −24 kN

Answer

V_x = −24 kN

6.Shear force & bending moment

Answer: M_x = 216 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·10 = 100 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 40·4 + 10·10·5

R_B = 660/10 = 66 kN

ΣF_y = 0: R_A = 140 − 66 = 74 kN

Cut at x = 4 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 = 74·4 − 10·4·2 = 216 kN·m

Answer

M_x = 216 kN·m

7.Centroid & second moment of area

Answer: r = 79.4 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 11259 mm²

I = π(D⁴ − d⁴)/64 = 7.098 × 10⁷ mm⁴

Radius of gyration

r = √(I/A) = √(7.098 × 10⁷/11259) = 79.4 mm

Answer

r = 79.4 mm

8.Centroid & second moment of area

Answer: M_allow = 3.56 kN·m

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.

Centroid

ȳ = (1200·205 + 1600·100)/2800 = 145 mm

top edge: c = 210 − 145 = 65 mm

Second moment of area

I = 120·10³/12 + 1200·60² + 8·200³/12 + 1600·45² = 1.29 × 10⁷ mm⁴

Moment allowed by each edge

Sagging: the top is in compression, the bottom in tension.

top edge: M = 90·1.29 × 10⁷/65 = 17.87 kN·m

bottom edge: M = 40·1.29 × 10⁷/145 = 3.56 kN·m

Governing value

M_allow = min = 3.56 kN·m (bottom edge)

Answer

M_allow = 3.56 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_abs = 55.7 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 = 75 MPa, R = √((−35)² + 10²) = 36.4 MPa

σ₁ = 111.4 MPa, σ₂ = 38.6 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(72.8, 111.4, 38.6)/2 = 55.7 MPa

Answer

τ_abs = 55.7 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: σ_n = 218.6 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. Thin-walled cylinder: hoop stress σ_h = pr/t and axial stress σ_a = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 1200/2 = 600 mm

σ_h = pr/t = 2·600/5 = 240 MPa

σ_a = pr/(2t) = 120 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 = 120·sin²25° + 240·cos²25° = 218.6 MPa

Answer

σ_n = 218.6 MPa

11.Beam deflection

Answer: δ_max = 14.49 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 = 4000 mm

EI = 4.6 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·20·4000⁴/(384·4.6 × 10¹²) = 14.49 mm

Answer

δ_max = 14.49 mm

12.Beam deflection

Answer: δ_max = 4.441 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 = 24000 N, w = 4 N/mm, L = 3000 mm, EI = 3.99 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 24000·3000³/(48·3.99 × 10¹²) = 3.383 mm

δ_w = 5wL⁴/(384EI) = 5·4·3000⁴/(384·3.99 × 10¹²) = 1.057 mm

Add them

δ_max = 3.383 + 1.057 = 4.441 mm

Answer

δ_max = 4.441 mm

13.Column buckling

Answer: P_cr = 14.1 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·40000/1400² = 14099 N = 14.1 kN

Answer

P_cr = 14.1 kN

14.Column buckling

Answer: d = 107.3 mm

Required critical load

P_cr = n·P = 3·85000 = 255000 N

KL = 0.7·6000 = 4200 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·255000·4200²/(π³·70000))^(1/4) = 107.3 mm

Answer

d = 107.3 mm

Check: the slenderness KL/r is 156.5, so the column is slender and Euler applies.

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