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 copper bar under the axial load.

P = 105 kNd = 25 mmL = 0.5 mE = 117 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 = 90 kNsegment 1: Ø30 mm, 1 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1 m, copper (E = 117 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 500 N·m. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

T = 500 N·md_i/d_o = 0.75τ_allow = 50 MPa

d_o = mm

4.Torsion of shaftsHard

A hollow shaft must transmit 50 kW at 1800 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

P = 50 kWN = 1800 rpmd_i/d_o = 0.75τ_allow = 40 MPa

d_o = 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.

15 kN @ 10 m20 kN @ 8 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 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 = 4 mP = 30 kN at 1 mw = 12 kN/m from 0 to 4 m

M_max = kN·m

7.Centroid & second moment of areaMedium

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 280 mmt_f = 25 mmt_w = 20 mmh_w = 260 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 = 140 mmt_f = 10 mmt_w = 14 mmh_w = 300 mmM = 15 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 130 MPa, σᵧ = 60 MPa and τₓᵧ = 20 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 = 130 MPaσ_y = 60 MPaτ_xy = 20 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 10 MPa, σᵧ = 60 MPa and τₓᵧ = −60 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 10 MPaσ_y = 60 MPaτ_xy = −60 MPaσ_Y = 350 MPa

n =

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 = 15 kNw = 20 kN/mL = 1.5 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the maximum deflection δ_max of the beam.

P = 6 kNb = 1.5 mL = 4 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_max = mm

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 = 2 mE = 70 GPa (aluminium)b × h = 25 × 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. 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: δ = 0.9141 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·25²/4 = 490.9 mm²

σ = P/A = 105·1000/490.9 = 213.9 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 213.9/117000 = 0.001828 = 1828 µε

Elongation

δ = ε·L = 0.001828·500 mm = 0.9141 mm

Answer

δ = 0.9141 mm

2.Axial stress & strain

Answer: σ_max = 286.5 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·30²/4 = 706.9 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 127.3 MPa

σ₂ = 286.5 MPa

Largest stress

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

σ_max = 286.5 MPa

Answer

σ_max = 286.5 MPa

3.Torsion of shafts

Answer: d_o = 42.08 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 = 500 N·m = 500000 N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·500000/(π·50·0.6836)) = 42.08 mm

d_i = 0.75·42.08 = 31.56 mm

Answer

d_o = 42.08 mm

A solid shaft would need d = 37.07 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d_o = 36.69 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 50000/188.5 = 265.3 N·m

Outer diameter from the allowable stress

T = 265.3 N·m = 265258 N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·265258/(π·40·0.6836)) = 36.69 mm

d_i = 0.75·36.69 = 27.52 mm

Answer

d_o = 36.69 mm

A solid shaft would need d = 32.32 mm; the hollow one is a little larger but much lighter.

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 = 9.167 kN, R₂ = 25.83 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 = 18.33, M_C = 36.67, M_D = 55, M_E = 73.33 kN·m, M_F = 51.67, M_G = 0

Answer

Mark E: M_E = 73.33 kN·m

6.Shear force & bending moment

Answer: M_max = 41.34 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 = 12·4 = 48 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 30·1 + 12·4·2

R_B = 126/4 = 31.5 kN

ΣF_y = 0: R_A = 78 − 31.5 = 46.5 kN

Where is V = 0?

R_A − P − w·x₀ = 0 → x₀ = (46.5 − 30)/12 = 1.375 m

Maximum moment

M_max = R_A·x₀ − P·(x₀ − 1) − w·x₀²/2 = 41.34 kN·m

Answer

M_max = 41.34 kN·m

It occurs at x = 1.375 m from A.

7.Centroid & second moment of area

Answer: I_x = 90.24 × 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 two rectangles

Measure every y from the bottom edge.

flange: A_f = 280·25 = 7000 mm², y_f = 260 + 25/2 = 272.5 mm

web: A_w = 20·260 = 5200 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (7000·272.5 + 5200·130)/12200 = 211.8 mm

Parallel-axis theorem

flange: 280·25³/12 + 7000·(272.5 − 211.8)² = 2.619 × 10⁷ mm⁴

web: 20·260³/12 + 5200·(130 − 211.8)² = 6.406 × 10⁷ mm⁴

I_x = 2.619 × 10⁷ + 6.406 × 10⁷ = 9.024 × 10⁷ mm⁴ = 90.24 × 10⁶ mm⁴

Answer

I_x = 90.24 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 49.9 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 140·10 = 1400 mm², y_f = 300 + 10/2 = 305 mm

web: A_w = 14·300 = 4200 mm², y_w = 300/2 = 150 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1400·305 + 4200·150)/5600 = 188.8 mm

Parallel-axis theorem

flange: 140·10³/12 + 1400·(305 − 188.8)² = 1.893 × 10⁷ mm⁴

web: 14·300³/12 + 4200·(150 − 188.8)² = 3.781 × 10⁷ mm⁴

I_x = 1.893 × 10⁷ + 3.781 × 10⁷ = 5.674 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 188.8 mm

σ = M·y/I = 1.5 × 10⁷·188.8/5.674 × 10⁷ = 49.9 MPa

Answer

σ_bottom = 49.9 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 14.87 °

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(2·20, 130 − 60) = 29.74°

Angle on the element

θ_p = 14.87°

Answer

θ_p = 14.87°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 2.692

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 = 65 MPa

σ₁ = 100 MPa, σ₂ = −30 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 130 MPa

n = 350/130 = 2.692

Answer

n = 2.692

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

11.Beam deflection

Answer: δ = 12.3 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 = 15000 N, w = 20 N/mm, L = 1500 mm

EI = 2.4 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 15000·1500³/(3·2.4 × 10¹²) = 7.031 mm

δ_w = wL⁴/(8EI) = 20·1500⁴/(8·2.4 × 10¹²) = 5.273 mm

Add them

δ = 7.031 + 5.273 = 12.3 mm

Answer

δ = 12.3 mm

12.Beam deflection

Answer: δ_max = 4.329 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 = 6000 N, b = 1500 mm, L = 4000 mm

EI = 1.7 × 10¹² N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((4000² − 1500²)/3) = 2141 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 4.329 mm

Answer

δ_max = 4.329 mm

13.Column buckling

Answer: P_cr = 17.99 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 = 80·25³/12 = 104167 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·104167/2000² = 17991 N = 17.99 kN

Answer

P_cr = 17.99 kN

14.Column buckling

Answer: σ_cr = 7.676 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 = 1·6000 = 6000 mm

KL/r = 6000/20 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa

Answer

σ_cr = 7.676 MPa

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