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 = 20 kNd = 20 mmL = 1 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 total elongation δ.

P = 10 kNsegment 1: Ø25 mm, 1 m, copper (E = 117 GPa)segment 2: Ø16 mm, 0.5 m, steel (E = 200 GPa)

δ = mm

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 = 200 N·md = 40 mmL = 0.5 mG = 26 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 = 75 mmd_i = 60 mm

τ_max = MPa

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.

45 kN @ 6 m8 kN/m, 2–8 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 over part of its span. Find the maximum bending moment M_max in the beam.

L = 4 mw = 10 kN/m from 0 to 2 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 = 180 mmt_f = 30 mmt_w = 18 mmh_w = 160 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 10 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 120 mmt_f = 30 mmt_w = 12 mmh_w = 220 mmM = 10 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 40 MPa, σᵧ = −70 MPa and τₓᵧ = 20 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 40 MPaσ_y = −70 MPaτ_xy = 20 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −50 MPa, σᵧ = 100 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.

σ_x = −50 MPaσ_y = 100 MPaτ_xy = 30 MPaθ = 45°

σ_θ = 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 = 8 kN/mL = 8 mE = 200 GPa (steel)I = 160 × 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 size of the bending moment M_A at the fixed end.

L = 6 mw = 12 kN/m

M_A = kN·m

13.Column bucklingMedium

A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 3 mE = 70 GPa (aluminium)d = 20 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 115 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 115 kNn = 2L = 3 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.5441 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·20²/4 = 314.2 mm²

σ = P/A = 20·1000/314.2 = 63.66 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 63.66/117000 = 0.0005441 = 544.1 µε

Elongation

δ = ε·L = 0.0005441·1000 mm = 0.5441 mm

Answer

δ = 0.5441 mm

2.Axial stress & strain

Answer: δ = 0.2985 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 = 20.37 MPa

σ₂ = 49.74 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 20.37·1000/117000 = 0.1741 mm

δ₂ = σ₂L₂/E₂ = 49.74·500/200000 = 0.1243 mm

δ = δ₁ + δ₂ = 0.2985 mm

Answer

δ = 0.2985 mm

3.Torsion of shafts

Answer: φ = 0.8768 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·500/(26000·251327) = 0.0153 rad

φ = 0.0153·180/π = 0.8768°

Answer

φ = 0.8768°

4.Torsion of shafts

Answer: τ_max = 13.02 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(75⁴ − 60⁴)/32 = 1.834 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 636620·37.5/1.834 × 10⁶ = 13.02 MPa

Answer

τ_max = 13.02 MPa

5.Shear force & bending moment

Answer: D (x = 6 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 = 50.5 kN, R₂ = 42.5 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 = 101, M_C = 186, M_D = 239 kN·m, M_E = 170, M_F = 85, M_G = 0

Answer

Mark D: M_D = 239 kN·m

6.Shear force & bending moment

Answer: M_max = 11.25 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 = 10·2 = 20 kN, acting at x̄ = 1 m

ΣM_A = 0: R_B·4 = 10·2·1

R_B = 20/4 = 5 kN

ΣF_y = 0: R_A = 20 − 5 = 15 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 − 0) = 0

x₀ = 0 + 15/10 = 1.5 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 0)²/2 = 15·1.5 − 10·1.5²/2 = 11.25 kN·m

Answer

M_max = 11.25 kN·m

7.Centroid & second moment of area

Answer: I_x = 23.5 × 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 = 180·30 = 5400 mm², y_f = 160 + 30/2 = 175 mm

web: A_w = 18·160 = 2880 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (5400·175 + 2880·80)/8280 = 142 mm

Parallel-axis theorem

flange: 180·30³/12 + 5400·(175 − 142)² = 6.301 × 10⁶ mm⁴

web: 18·160³/12 + 2880·(80 − 142)² = 1.72 × 10⁷ mm⁴

I_x = 6.301 × 10⁶ + 1.72 × 10⁷ = 2.35 × 10⁷ mm⁴ = 23.5 × 10⁶ mm⁴

Answer

I_x = 23.5 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 19.55 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 120·30 = 3600 mm², y_f = 220 + 30/2 = 235 mm

web: A_w = 12·220 = 2640 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3600·235 + 2640·110)/6240 = 182.1 mm

Parallel-axis theorem

flange: 120·30³/12 + 3600·(235 − 182.1)² = 1.034 × 10⁷ mm⁴

web: 12·220³/12 + 2640·(110 − 182.1)² = 2.438 × 10⁷ mm⁴

I_x = 1.034 × 10⁷ + 2.438 × 10⁷ = 3.472 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 250 − 182.1 = 67.88 mm

σ = M·y/I = 1 × 10⁷·67.88/3.472 × 10⁷ = 19.55 MPa

Answer

σ_top = 19.55 MPa

9.Mohr’s circle & stress transformation

Answer: θ_s = −35.01 °

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

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = 20/58.52 = 0.3417, sin 2θ_s = −55/58.52 = −0.9398

2θ_s = atan2(−55, 20) = −70.02°

Angle on the element

θ_s = −35.01°

check: θ_s − θ_p = −35.01 − 9.992 = −45°

Answer

θ_s = −35.01°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 55 MPa

Transformation equation

σ_θ = 25 + −75·0 + 30·1 = 55 MPa

Answer

σ_θ = 55 MPa

11.Beam deflection

Answer: δ_max = 13.33 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 = 8 N/mm, L = 8000 mm

EI = 3.2 × 10¹³ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·8000⁴/(384·3.2 × 10¹³) = 13.33 mm

Answer

δ_max = 13.33 mm

12.Beam deflection

Answer: M_A = 54 kN·m

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 = 3·12·6/8 = 27 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = wL²/2 − R_B·L = 216 − 162 = 54 kN·m

Answer

M_A = 54 kN·m

13.Column buckling

Answer: P_cr = 2.412 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·20⁴/64 = 7854 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·7854/1500² = 2412 N = 2.412 kN

Answer

P_cr = 2.412 kN

14.Column buckling

Answer: d = 62.5 mm

Required critical load

P_cr = n·P = 2·115000 = 230000 N

KL = 0.5·3000 = 1500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·230000·1500²/(π³·70000))^(1/4) = 62.5 mm

Answer

d = 62.5 mm

Check: the slenderness KL/r is 96, so the column is slender and Euler applies.

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