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

A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 50 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: copper, Ø50 mm, 500 mm, E = 120 GPaδ = 0.3 mm

P = kN

2.Axial stress & strainHard

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

P = 40 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, steel (E = 200 GPa)

δ = mm

3.Torsion of shaftsMedium

A hollow steel shaft (outer diameter 40 mm, inner diameter 20 mm, length 2 m) is fixed at one end and twisted by a torque of 1000 N·m at the other. Find the angle of twist φ in degrees.

T = 1000 N·md_o = 40 mmd_i = 20 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 1450 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 100 kWN = 1450 rpmτ_allow = 80 MPa

d = 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.

50 kN @ 5 m20 kN @ 1 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 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 = 5 mP = 25 kN at 1.5 mw = 6 kN/m from 0 to 5 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 = 140 mmt_f = 25 mmt_w = 12 mmh_w = 200 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 40 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 240 mmt_f = 25 mmt_w = 18 mmh_w = 240 mmM = 40 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −30 MPa, σᵧ = −80 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −30 MPaσ_y = −80 MPaτ_xy = −20 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 50 mm carries an axial tensile force of 80 kN together with a torque of 500 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 50 mmF = 80 kNT = 500 N·m

σ₁ = 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 = 4 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 27 kNw = 20 kN/mL = 2.5 mE = 70 GPa (aluminium)I = 84 × 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 = 1.5 mE = 70 GPa (aluminium)d = 25 mm

P_cr = kN

14.Column bucklingHard

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

L = 2.5 mE = 70 GPa (aluminium)d = 25 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 57.94 kN

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

A₁ = 490.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(120000·1963) = 2.122 × 10⁻⁶ mm/N

Load

P = 0.3/5.178 × 10⁻⁶ = 57939 N = 57.94 kN

Answer

P = 57.94 kN

2.Axial stress & strain

Answer: δ = 1.112 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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 31.83 MPa

σ₂ = 353.7 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 31.83·500/70000 = 0.2274 mm

δ₂ = σ₂L₂/E₂ = 353.7·500/200000 = 0.8842 mm

δ = δ₁ + δ₂ = 1.112 mm

Answer

δ = 1.112 mm

3.Torsion of shafts

Answer: φ = 6.079 °

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 − 20⁴)/32 = 235619 mm⁴

Angle of twist

T = 1000 N·m = 1 × 10⁶ N·mm

L = 2 m = 2000 mm

G = 80 GPa = 80000 MPa

φ = TL/(GJ) = 1 × 10⁶·2000/(80000·235619) = 0.1061 rad

φ = 0.1061·180/π = 6.079°

Answer

φ = 6.079°

4.Torsion of shafts

Answer: d = 34.74 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·80)) = 34.74 mm

Answer

d = 34.74 mm

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

5.Shear force & bending moment

Answer: F (x = 5 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 = 25 kN, R₂ = 45 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 = 25, M_C = 30, M_D = 35, M_E = 40, M_F = 45 kN·m, M_G = 0

Answer

Mark F: M_F = 45 kN·m

6.Shear force & bending moment

Answer: M_max = 42 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 = 6·5 = 30 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 25·1.5 + 6·5·2.5

R_B = 112.5/5 = 22.5 kN

ΣF_y = 0: R_A = 55 − 22.5 = 32.5 kN

Where is V = 0?

V(1.5⁻) = R_A − w·1.5 = 23.5 kN

V(1.5⁺) = 23.5 − 25 = −1.5 kN

The shear changes sign at the load, so the maximum is under the load.

Maximum moment

M_max = R_A·1.5 − w·1.5²/2 = 32.5·1.5 − 6·1.5²/2 = 42 kN·m

Answer

M_max = 42 kN·m

It occurs at x = 1.5 m from A.

7.Centroid & second moment of area

Answer: I_x = 26.2 × 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 = 140·25 = 3500 mm², y_f = 200 + 25/2 = 212.5 mm

web: A_w = 12·200 = 2400 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3500·212.5 + 2400·100)/5900 = 166.7 mm

Parallel-axis theorem

flange: 140·25³/12 + 3500·(212.5 − 166.7)² = 7.512 × 10⁶ mm⁴

web: 12·200³/12 + 2400·(100 − 166.7)² = 1.869 × 10⁷ mm⁴

I_x = 7.512 × 10⁶ + 1.869 × 10⁷ = 2.62 × 10⁷ mm⁴ = 26.2 × 10⁶ mm⁴

Answer

I_x = 26.2 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 41.73 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 240·25 = 6000 mm², y_f = 240 + 25/2 = 252.5 mm

web: A_w = 18·240 = 4320 mm², y_w = 240/2 = 120 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·252.5 + 4320·120)/10320 = 197 mm

Parallel-axis theorem

flange: 240·25³/12 + 6000·(252.5 − 197)² = 1.877 × 10⁷ mm⁴

web: 18·240³/12 + 4320·(120 − 197)² = 4.637 × 10⁷ mm⁴

I_x = 1.877 × 10⁷ + 4.637 × 10⁷ = 6.514 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 265 − 197 = 67.97 mm

σ = M·y/I = 4 × 10⁷·67.97/6.514 × 10⁷ = 41.73 MPa

Answer

σ_top = 41.73 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 32.02 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.

Radius of Mohr’s circle

τ_max = R = √(((−30 − −80)/2)² + −20²) = 32.02 MPa

Answer

τ_max = 32.02 MPa

10.Mohr’s circle & stress transformation

Answer: σ₁ = 49.18 MPa

Stresses at the surface

σ = F/A = 80000/(π·50²/4) = 80000/1963 = 40.74 MPa

τ = 16T/(πd³) = 16·500000/(π·50³) = 20.37 MPa

Mohr’s circle

C = σ/2 = 20.37 MPa

R = √(20.37² + 20.37²) = 28.81 MPa

σ₁ = C + R = 49.18 MPa

σ₂ = C − R = −8.438 MPa

Answer

σ₁ = 49.18 MPa

11.Beam deflection

Answer: δ_max = 28.86 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 = 4000 mm

EI = 9.24 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·4000⁴/(384·9.24 × 10¹¹) = 28.86 mm

Answer

δ_max = 28.86 mm

12.Beam deflection

Answer: δ_max = 3.225 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 = 27000 N, w = 20 N/mm, L = 2500 mm, EI = 5.88 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 27000·2500³/(48·5.88 × 10¹²) = 1.495 mm

δ_w = 5wL⁴/(384EI) = 5·20·2500⁴/(384·5.88 × 10¹²) = 1.73 mm

Add them

δ_max = 1.495 + 1.73 = 3.225 mm

Answer

δ_max = 3.225 mm

13.Column buckling

Answer: P_cr = 5.888 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 = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·19175/1500² = 5888 N = 5.888 kN

Answer

P_cr = 5.888 kN

14.Column buckling

Answer: σ_cr = 8.812 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 0.7·2500 = 1750 mm

KL/r = 1750/6.25 = 280

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/280² = 8.812 MPa

Answer

σ_cr = 8.812 MPa

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