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 bar is made of a steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: brass, Ø40 mm, 400 mm, E = 100 GPaδ = 0.4 mm

P = kN

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 = 90 kNsegment 1: Ø40 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 2 m, steel (E = 200 GPa)

σ_max = 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 = 250 N·md = 50 mmL = 2 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft transmits 10 kW at 1450 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 10 kWN = 1450 rpmτ_allow = 60 MPaθ_allow = 0.75°/mG = 26 GPa

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.

25 kN @ 2 m15 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

Find the bending moment M_x at the section x = 5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 40 kN at 5 mw = 10 kN/m from 0 to 8 mx = 5 m

M_x = 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 = 100 mmt_f = 15 mmt_w = 16 mmh_w = 160 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A symmetric I-section has flanges 190 mm wide and 14 mm thick, an overall depth of 360 mm and a web 10 mm thick. It carries a bending moment of 200 kN·m about the axis x. Find the size of the maximum bending stress σ_max in the section.

B = 190 mmH = 360 mmt_f = 14 mmt_w = 10 mmM = 200 kN·m

σ_max = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 10 MPa, σᵧ = 80 MPa and τₓᵧ = 50 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 = 10 MPaσ_y = 80 MPaτ_xy = 50 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 110 MPa, σᵧ = 90 MPa and τₓᵧ = −60 MPa (σ_z = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 110 MPaσ_y = 90 MPaτ_xy = −60 MPaσ_Y = 275 MPa

n =

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.

P = 10 kNw = 12 kN/mL = 3 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 2 m from support B. Find the maximum deflection δ_max of the beam.

P = 14 kNb = 2 mL = 5 mE = 11 GPa (timber)I = 84 × 10⁶ mm⁴

δ_max = mm

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 = 2.5 mE = 200 GPa (steel)d = 40 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. 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: P = 75.4 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 = 706.9 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(100000·1257) = 3.183 × 10⁻⁶ mm/N

Load

P = 0.4/5.305 × 10⁻⁶ = 75398 N = 75.4 kN

Answer

P = 75.4 kN

2.Axial stress & strain

Answer: σ_max = 447.6 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·16²/4 = 201.1 mm²

σ₁ = 71.62 MPa

σ₂ = 447.6 MPa

Largest stress

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

σ_max = 447.6 MPa

Answer

σ_max = 447.6 MPa

3.Torsion of shafts

Answer: φ = 1.197 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·2000/(39000·613592) = 0.02089 rad

φ = 0.02089·180/π = 1.197°

Answer

φ = 1.197°

4.Torsion of shafts

Answer: d = 37.47 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 10000/151.8 = 65.86 N·m

Diameter for strength

T = 65.86 N·m = 65857 N·mm

d_τ = ∛(16·65857/(π·60)) = 17.75 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm

G = 26 GPa = 26000 MPa

d_φ = (32·65857/(π·26000·1.309 × 10⁻⁵))^1/4 = 37.47 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(17.75, 37.47) = 37.47 mm

Answer

d = 37.47 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: C (x = 2 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 = 29.17 kN, R₂ = 10.83 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 = 29.17, M_C = 43.33 kN·m, M_D = 32.5, M_E = 21.67, M_F = 10.83, M_G = 0

Answer

Mark C: M_C = 43.33 kN·m

6.Shear force & bending moment

Answer: M_x = 150 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 = 10·8 = 80 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 40·5 + 10·8·4

R_B = 520/8 = 65 kN

ΣF_y = 0: R_A = 120 − 65 = 55 kN

Cut at x = 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 = 55·5 − 10·5·2.5 = 150 kN·m

Answer

M_x = 150 kN·m

7.Centroid & second moment of area

Answer: I_x = 12.73 × 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 = 100·15 = 1500 mm², y_f = 160 + 15/2 = 167.5 mm

web: A_w = 16·160 = 2560 mm², y_w = 160/2 = 80 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1500·167.5 + 2560·80)/4060 = 112.3 mm

Parallel-axis theorem

flange: 100·15³/12 + 1500·(167.5 − 112.3)² = 4.594 × 10⁶ mm⁴

web: 16·160³/12 + 2560·(80 − 112.3)² = 8.137 × 10⁶ mm⁴

I_x = 4.594 × 10⁶ + 8.137 × 10⁶ = 1.273 × 10⁷ mm⁴ = 12.73 × 10⁶ mm⁴

Answer

I_x = 12.73 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_max = 189.7 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 = 360 − 2·14 = 332 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (190·360³ − 180·332³)/12

I_x = 1.898 × 10⁸ mm⁴

Bending stress

y = 180 mm

σ = M·y/I = 2 × 10⁸·180/1.898 × 10⁸ = 189.7 MPa

Answer

σ_max = 189.7 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 62.5 °

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(2·50, 10 − 80) = 125°

Angle on the element

θ_p = 62.5°

Answer

θ_p = 62.5°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: n = 1.71

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 = 100 MPa, R = 60.83 MPa

σ₁ = 160.8 MPa, σ₂ = 39.17 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 160.8 MPa

n = 275/160.8 = 1.71

Answer

n = 1.71

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

11.Beam deflection

Answer: δ = 53.01 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, w = 12 N/mm, L = 3000 mm

EI = 3.99 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 10000·3000³/(3·3.99 × 10¹²) = 22.56 mm

δ_w = wL⁴/(8EI) = 12·3000⁴/(8·3.99 × 10¹²) = 30.45 mm

Add them

δ = 22.56 + 30.45 = 53.01 mm

Answer

δ = 53.01 mm

12.Beam deflection

Answer: δ_max = 37.41 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 = 14000 N, b = 2000 mm, L = 5000 mm

EI = 9.24 × 10¹¹ N·mm²

Position of the maximum

Setting the slope in the longer segment to zero gives the position.

x = √((L² − b²)/3) = √((5000² − 2000²)/3) = 2646 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 37.41 mm

Answer

δ_max = 37.41 mm

13.Column buckling

Answer: P_cr = 81 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·125664/1750² = 80996 N = 81 kN

Answer

P_cr = 81 kN

14.Column buckling

Answer: σ_cr = 30.71 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 = 1·1500 = 1500 mm

KL/r = 1500/10 = 150

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/150² = 30.71 MPa

Answer

σ_cr = 30.71 MPa

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