Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 = 95 kNd = 30 mmL = 1 mE = 100 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 = 10 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 2 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 = 100 N·md = 20 mmL = 1.2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 1450 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 100 kWN = 1450 rpmτ_allow = 75 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 6 mw = 8 kN/m from 0 to 6 mx = 1.5 m

M_x = kN·m

6.Shear force & bending momentHard

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.

35 kN @ 0 m55 kN @ 12 m6 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)

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 200 mm wide with two upright legs, 200 mm high overall. Base and legs are all 8 mm thick. Find the height ȳ of the centroid above the bottom edge.

b = 200 mmh = 200 mmt = 8 mm

ȳ = 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 100 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 100 mmt_f = 15 mmt_w = 20 mmh_w = 280 mmσ_t,allow = 40 MPaσ_c,allow = 100 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 70 MPa and σᵧ = −50 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 125 MPa?

σ_x = 70 MPaσ_y = −50 MPaσ₁ = 125 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = 40 MPa and τₓᵧ = 20 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = −70 MPaσ_y = 40 MPaτ_xy = 20 MPa

σ_vM = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1 m from A. Find the deflection δ_P under the load.

P = 10 kNa = 1 mL = 2 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the reaction R_B at the prop.

L = 8 mP = 15 kN at 4 m

R_B = kN

13.Column bucklingMedium

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

L = 5 mE = 200 GPa (steel)b × h = 20 × 60 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1.5 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 1.344 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·30²/4 = 706.9 mm²

σ = P/A = 95·1000/706.9 = 134.4 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 134.4/100000 = 0.001344 = 1344 µε

Elongation

δ = ε·L = 0.001344·1000 mm = 1.344 mm

Answer

δ = 1.344 mm

2.Axial stress & strain

Answer: σ_max = 31.83 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·40²/4 = 1257 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 7.958 MPa

σ₂ = 31.83 MPa

Largest stress

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

σ_max = 31.83 MPa

Answer

σ_max = 31.83 MPa

3.Torsion of shafts

Answer: φ = 16.83 °

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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1200/(26000·15708) = 0.2938 rad

φ = 0.2938·180/π = 16.83°

Answer

φ = 16.83°

4.Torsion of shafts

Answer: d = 35.5 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 100000/151.8 = 658.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·658572/(π·75)) = 35.5 mm

Answer

d = 35.5 mm

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

5.Shear force & bending moment

Answer: M_x = 27 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·6 = 48 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·6 = 8·6·3

R_B = 144/6 = 24 kN

ΣF_y = 0: R_A = 48 − 24 = 24 kN

Cut at x = 1.5 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 = 24·1.5 − 8·1.5·0.75 = 27 kN·m

Answer

M_x = 27 kN·m

6.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 = 28 kN, R₂ = 98 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 = −70, M_C = −96, M_D = −146, M_E = −220 kN·m, M_F = −110, M_G = 0

Answer

Mark E: M_E = −220 kN·m

7.Centroid & second moment of area

Answer: ȳ = 69.75 mm

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 = 200·8 = 1600 mm², y₁ = 4 mm

two legs: A₂ = 2·8·192 = 3072 mm², y₂ = 8 + 192/2 = 104 mm

Centroid

ȳ = (1600·4 + 3072·104)/4672 = 69.75 mm

Answer

ȳ = 69.75 mm

8.Centroid & second moment of area

Answer: M_allow = 14.57 kN·m

Centroid

ȳ = (1500·287.5 + 5600·140)/7100 = 171.2 mm

top edge: c = 295 − 171.2 = 123.8 mm

Second moment of area

I = 100·15³/12 + 1500·116.3² + 20·280³/12 + 5600·31.16² = 6.235 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 100·6.235 × 10⁷/123.8 = 50.35 kN·m

bottom edge: M = 40·6.235 × 10⁷/171.2 = 14.57 kN·m

Governing value

M_allow = min = 14.57 kN·m (bottom edge)

Answer

M_allow = 14.57 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_xy = 98.11 MPa

Required radius

C = (70 + (−50))/2 = 10 MPa

R = σ₁ − C = 125 − 10 = 115 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(115² − 60²) = 98.11 MPa

Answer

τ_xy = 98.11 MPa

10.Mohr’s circle & stress transformation

Answer: σ_vM = 102.5 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. 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 = −15 MPa, R = 58.52 MPa

σ₁ = 43.52 MPa, σ₂ = −73.52 MPa, σ₃ = 0

Von Mises

σ_vM = √(43.52² − 43.52·(−73.52) + (−73.52)²) = 102.5 MPa

Answer

σ_vM = 102.5 MPa

11.Beam deflection

Answer: δ_P = 1.961 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 = 10000 N, a = 1000 mm, L = 2000 mm

EI = 1.7 × 10¹² N·mm²

Deflection under the load

δ_P = Pa³/(3EI) = 10000·1000³/(3·1.7 × 10¹²) = 1.961 mm

Answer

δ_P = 1.961 mm

12.Beam deflection

Answer: R_B = 4.688 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

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) = 5PL³/(48EI) → R_B = 5P/16 = 5·15/16 = 4.688 kN

Answer

R_B = 4.688 kN

13.Column buckling

Answer: P_cr = 12.63 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 = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·40000/2500² = 12633 N = 12.63 kN

Answer

P_cr = 12.63 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·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 2·1500 = 3000 mm

KL/r = 3000/10 = 300

Critical stress

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

Answer

σ_cr = 7.676 MPa

ExamProblems · examproblems.com/courses/mechanics-of-materials