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 = 105 kNd = 20 mmL = 1.5 mE = 100 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 500 mm, E = 110 GPaP = 40 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid brass 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 = 750 N·md = 40 mmL = 1.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 10 kW at 3000 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 10 kWN = 3000 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 60 kN·m at x = 1 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 4 mM₀ = 60 kN·m clockwise, at 1 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mP = 45 kN at 3 mw = 6 kN/m from 0 to 5 mx = 3 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 260 mm wide with two upright legs, 100 mm high overall. Base and legs are all 12 mm thick. Find its second moment of area I_x about the horizontal axis x through its centroid.

b = 260 mmh = 100 mmt = 12 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 110 mm wide and 14 mm thick, an overall depth of 380 mm and a web 6 mm thick. It carries a bending moment of 30 kN·m about the axis x. Find the size of the bending stress σ_j at the junction of the web and the top flange.

B = 110 mmH = 380 mmt_f = 14 mmt_w = 6 mmM = 30 kN·m

σ_j = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −60 MPa, σᵧ = −70 MPa and τₓᵧ = −10 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 = −60 MPaσ_y = −70 MPaτ_xy = −10 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 70 MPa, σᵧ = 20 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 30° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 70 MPaσ_y = 20 MPaτ_xy = 30 MPaθ = -30°

τ_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 = 3 mE = 70 GPa (aluminium)I = 36 × 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 reaction R_B at the prop.

L = 6 mw = 5 kN/m

R_B = kN

13.Column bucklingMedium

A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 5 mE = 200 GPa (steel)d = 25 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 25 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 25 kNn = 2.5L = 5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 5.013 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 = 105·1000/314.2 = 334.2 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 334.2/100000 = 0.003342 = 3342 µε

Elongation

δ = ε·L = 0.003342·1500 mm = 5.013 mm

Answer

δ = 5.013 mm

2.Axial stress & strain

Answer: σ₂ = 11.23 MPa

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.

Stiffnesses

k₁ = 200000·1257/400 = 628319 N/mm

k₂ = 110000·706.9/500 = 155509 N/mm

Share of segment 2

F₂ = 40000·155509/783827 = 7936 N

Stress

σ₂ = 7936/706.9 = 11.23 MPa (compression)

Answer

σ₂ = 11.23 MPa

3.Torsion of shafts

Answer: φ = 6.576 °

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) = 750·1000·1500/(39000·251327) = 0.1148 rad

φ = 0.1148·180/π = 6.576°

Answer

φ = 6.576°

4.Torsion of shafts

Answer: d = 15.94 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 10000/314.2 = 31.83 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·31831/(π·40)) = 15.94 mm

Answer

d = 15.94 mm

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

5.Shear force & bending moment

Answer: M_x = 45 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 60/4 = 15 kN (upward)

R_A = −R_B = −15 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −15·1 + 60 = 45 kN·m

Answer

M_x = 45 kN·m

6.Shear force & bending moment

Answer: M_x = 72 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·5 = 30 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 45·3 + 6·5·2.5

R_B = 210/5 = 42 kN

ΣF_y = 0: R_A = 75 − 42 = 33 kN

Cut at x = 3 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 = 33·3 − 6·3·1.5 = 72 kN·m

Answer

M_x = 72 kN·m

7.Centroid & second moment of area

Answer: I_x = 4.549 × 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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 260·12 = 3120 mm², y₁ = 6 mm

two legs: A₂ = 2·12·88 = 2112 mm², y₂ = 12 + 88/2 = 56 mm

Centroid

ȳ = (3120·6 + 2112·56)/5232 = 26.18 mm

Parallel-axis theorem

base: 260·12³/12 + 3120·(6 − 26.18)² = 1.308 × 10⁶ mm⁴

two legs: 2·12·88³/12 + 2112·(56 − 26.18)² = 3.241 × 10⁶ mm⁴

I_x = 1.308 × 10⁶ + 3.241 × 10⁶

I_x = 4.549 × 10⁶ mm⁴ = 4.549 × 10⁶ mm⁴

Answer

I_x = 4.549 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_j = 42.24 MPa

Outer rectangle minus the two side cut-outs

Both cut-outs together are (B − t_w) wide and h_w = H − 2t_f high, and share the axis x.

h_w = 380 − 2·14 = 352 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (110·380³ − 104·352³)/12

I_x = 1.25 × 10⁸ mm⁴

Bending stress

y = 190 − 14 = 176 mm

σ = M·y/I = 3 × 10⁷·176/1.25 × 10⁸ = 42.24 MPa

Answer

σ_j = 42.24 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −31.72 °

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·−10, −60 − −70) = −63.43°

Angle on the element

θ_p = −31.72°

Answer

θ_p = −31.72°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: τ_θ = 36.65 MPa

Transformation equation

τ_θ = −25·−0.866 + 30·0.5 = 36.65 MPa

Answer

τ_θ = 36.65 MPa

11.Beam deflection

Answer: δ_max = 3.348 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 = 3000 mm

EI = 2.52 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·3000⁴/(384·2.52 × 10¹²) = 3.348 mm

Answer

δ_max = 3.348 mm

12.Beam deflection

Answer: R_B = 11.25 kN

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 = 3·5·6/8 = 11.25 kN

Answer

R_B = 11.25 kN

13.Column buckling

Answer: P_cr = 3.09 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 0.7·5000 = 3500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·19175/3500² = 3090 N = 3.09 kN

Answer

P_cr = 3.09 kN

14.Column buckling

Answer: d = 68.93 mm

Required critical load

P_cr = n·P = 2.5·25000 = 62500 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·62500·3500²/(π³·70000))^(1/4) = 68.93 mm

Answer

d = 68.93 mm

Check: the slenderness KL/r is 203.1, so the column is slender and Euler applies.

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