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 40 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 60 mm, E = 120 GPa). Both are 400 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 120 GPaL = 400 mm

δ = mm

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 30 kNsegment 1: Ø25 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 2 m, brass (E = 100 GPa)

δ = mm

3.Torsion of shaftsMedium

A solid steel shaft 1.5 m long carries a torque of 2000 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.

T = 2000 N·mL = 1.5 mG = 80 GPaφ_allow = 1°

d = mm

4.Torsion of shaftsHard

A hollow shaft must transmit 50 kW at 1000 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

P = 50 kWN = 1000 rpmd_i/d_o = 0.75τ_allow = 75 MPa

d_o = mm

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 2 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 = 50 kN at 2 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 over part of its span. Find the distance x₀ from A to the section where the bending moment is largest.

L = 5 mw = 4 kN/m from 1 to 3 m

x₀ = m

7.Centroid & second moment of areaMedium

A 200 mm × 240 mm plate has a circular hole of diameter 30 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 = 30 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 20 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 100 mmt_f = 25 mmt_w = 14 mmh_w = 100 mmM = 20 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = −20 MPa, σᵧ = 10 MPa and τₓᵧ = −40 MPa (tension positive). The element is rotated 35° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = −20 MPaσ_y = 10 MPaτ_xy = −40 MPaθ = 35°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −20 MPa, σᵧ = −60 MPa and τₓᵧ = 20 MPa (σ_z = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −20 MPaσ_y = −60 MPaτ_xy = 20 MPaσ_Y = 275 MPa

n =

11.Beam deflectionMedium

A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θ_A of the beam at support A, in radians.

w = 8 kN/mL = 6 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

θ_A = rad

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 8 kN/mL = 4 mE = 11 GPa (timber)

I_req = × 10⁶ mm⁴

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 × 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 140 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 140 kNn = 2L = 5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.03882 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·40²/4 = 1257 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·2199 = 5.152 × 10⁸ N

Shortening

δ = 50000·400/5.152 × 10⁸ = 0.03882 mm

Answer

δ = 0.03882 mm

2.Axial stress & strain

Answer: δ = 4.73 mm

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·16²/4 = 201.1 mm²

σ₁ = 61.12 MPa

σ₂ = 149.2 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 61.12·2000/70000 = 1.746 mm

δ₂ = σ₂L₂/E₂ = 149.2·2000/100000 = 2.984 mm

δ = δ₁ + δ₂ = 4.73 mm

Answer

δ = 4.73 mm

3.Torsion of shafts

Answer: d = 68.4 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.

Required polar moment

T = 2000 N·m = 2 × 10⁶ N·mm

L = 1.5 m = 1500 mm

G = 80 GPa = 80000 MPa

φ = 1·π/180 = 0.01745 rad

J = TL/(Gφ) = 2 × 10⁶·1500/(80000·0.01745) = 2.149 × 10⁶ mm⁴

Diameter

d = (32J/π)^1/4 = 68.4 mm

Answer

d = 68.4 mm

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

4.Torsion of shafts

Answer: d_o = 36.2 mm

Torque

ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s

T = P/ω = 50000/104.7 = 477.5 N·m

Outer diameter from the allowable stress

T = 477.5 N·m = 477465 N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·477465/(π·75·0.6836)) = 36.2 mm

d_i = 0.75·36.2 = 27.15 mm

Answer

d_o = 36.2 mm

A solid shaft would need d = 31.89 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: M_A = −109 kN·m

Keep the free-end part

P: −50·2 = −100 kN·m

w: −2·3²/2 = −9 kN·m

M_A = −100 − 9 = −109 kN·m

Answer

M_A = −109 kN·m

Negative means hogging: tension on the top face.

6.Shear force & bending moment

Answer: x₀ = 2.2 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 = 4·2 = 8 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·5 = 4·2·2

R_B = 16/5 = 3.2 kN

ΣF_y = 0: R_A = 8 − 3.2 = 4.8 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 1) = 0

x₀ = 1 + 4.8/4 = 2.2 m

Answer

x₀ = 2.2 m

7.Centroid & second moment of area

Answer: I_x = 228.7 × 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 − 706.9·72)/(48000 − 706.9) = 120.7 mm

Full plate about ȳ

200·240³/12 + 48000·−0.7174² = 230.4 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−48.72² = 1.717 × 10⁶ mm⁴

Subtract

I_x = 228.7 × 10⁶ mm⁴

Answer

I_x = 228.7 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 375.1 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 = 100·25 = 2500 mm², y_f = 100 + 25/2 = 112.5 mm

web: A_w = 14·100 = 1400 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2500·112.5 + 1400·50)/3900 = 90.06 mm

Parallel-axis theorem

flange: 100·25³/12 + 2500·(112.5 − 90.06)² = 1.389 × 10⁶ mm⁴

web: 14·100³/12 + 1400·(50 − 90.06)² = 3.414 × 10⁶ mm⁴

I_x = 1.389 × 10⁶ + 3.414 × 10⁶ = 4.802 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 90.06 mm

σ = M·y/I = 2 × 10⁷·90.06/4.802 × 10⁶ = 375.1 MPa

Answer

σ_bottom = 375.1 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 37.72 MPa

Transformation equation for the y′ face

σ_y′ = −5 − (−15)·0.342 − (−40)·0.9397 = 37.72 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = −47.72 MPa, σ_x′ + σ_y′ = −10 = σ_x + σ_y

Answer

σ_y′ = 37.72 MPa

10.Mohr’s circle & stress transformation

Answer: n = 4.348

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 = −40 MPa, R = 28.28 MPa

σ₁ = −11.72 MPa, σ₂ = −68.28 MPa, σ₃ = 0

Von Mises

σ_vM = √((−11.72)² − (−11.72)·(−68.28) + (−68.28)²) = 63.25 MPa

n = 275/63.25 = 4.348

Answer

n = 4.348

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

11.Beam deflection

Answer: θ_A = 0.01565 rad

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). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.

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 = 8 N/mm, L = 6000 mm

EI = 4.6 × 10¹² N·mm²

End slope

θ_A = wL³/(24EI) = 8·6000³/(24·4.6 × 10¹²) = 0.01565 rad

Answer

θ_A = 0.01565 rad

That is 0.8968°.

12.Beam deflection

Answer: I_req = 218.2 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 4000/360 = 11.11 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·8·4000⁴/(384·11000·11.11) = 2.182 × 10⁸ mm⁴

I_req = 218.2 × 10⁶ mm⁴

Answer

I_req = 218.2 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 8.996 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 = 40·25³/12 = 52083 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·52083/2000² = 8996 N = 8.996 kN

Answer

P_cr = 8.996 kN

14.Column buckling

Answer: d = 65.19 mm

Required critical load

P_cr = n·P = 2·140000 = 280000 N

KL = 0.5·5000 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·280000·2500²/(π³·200000))^(1/4) = 65.19 mm

Answer

d = 65.19 mm

Check: the slenderness KL/r is 153.4, so the column is slender and Euler applies.

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