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 axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).

P = 80 kNd = 30 mmL = 2.5 mE = 70 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 70 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø100/Ø70 mm, E = 120 GPa, σ_allow = 60 MPa

P_max = kN

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 = 400 N·md = 50 mmL = 2.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 3000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 3000 rpmd_o = 50 mmd_i = 25 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.

15 kN @ 10 m45 kN @ 6 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 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.

10 kN @ 0 m35 kN @ 6 m35 kN @ 3 m6 kN/m, 1–5 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)

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 = 200 mmt_f = 30 mmt_w = 10 mmh_w = 280 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 30 MPa in tension and 120 MPa in compression. Find the largest hogging bending moment M_allow the beam can carry (give its size).

b_f = 220 mmt_f = 30 mmt_w = 10 mmh_w = 100 mmσ_t,allow = 30 MPaσ_c,allow = 120 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 80 MPa, σᵧ = −80 MPa and τₓᵧ = 20 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 = 80 MPaσ_y = −80 MPaτ_xy = 20 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 20 MPa, σᵧ = 0 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.

σ_x = 20 MPaσ_y = 0 MPaτ_xy = 10 MPaθ = 30°

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 = 2 kN/mL = 6 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth h_req of the section.

w = 4 kN/mL = 4.5 mE = 11 GPa (timber)b = 100 mm

h_req = mm

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 × 50 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 90 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 90 kNn = 2.5L = 1.5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 1617 µε

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 = 80·1000/706.9 = 113.2 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 113.2/70000 = 0.001617 = 1617 µε

Answer

ε = 1617 µε

2.Axial stress & strain

Answer: P_max = 366 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

A_r = π·40²/4 = 1257 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·4006 = 7.32 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0005) = 0.0005 (copper)

Largest load

P_max = 0.0005·7.32 × 10⁸ = 365996 N = 366 kN

Answer

P_max = 366 kN

3.Torsion of shafts

Answer: φ = 2.394 °

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) = 400·1000·2500/(39000·613592) = 0.04179 rad

φ = 0.04179·180/π = 2.394°

Answer

φ = 2.394°

4.Torsion of shafts

Answer: τ_max = 10.38 MPa

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 75000/314.2 = 238.7 N·m

Polar second moment of area

J = π(50⁴ − 25⁴)/32 = 575243 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 238732·25/575243 = 10.38 MPa

Answer

τ_max = 10.38 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 = 25 kN, R₂ = 35 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 = 50, M_C = 100, M_D = 150 kN·m, M_E = 110, M_F = 70, M_G = 0

Answer

Mark D: M_D = 150 kN·m

6.Shear force & bending moment

Answer: F (x = 5 m)

Support reactions

Take moments about one support, then use the vertical force balance.

R₁ = 33.25 kN, R₂ = 70.75 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 = −10, M_C = 10.25, M_D = 24.5, M_E = −2.25, M_F = −35 kN·m, M_G = 0

Answer

Mark F: M_F = −35 kN·m

7.Centroid & second moment of area

Answer: I_x = 64.61 × 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 = 200·30 = 6000 mm², y_f = 280 + 30/2 = 295 mm

web: A_w = 10·280 = 2800 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·295 + 2800·140)/8800 = 245.7 mm

Parallel-axis theorem

flange: 200·30³/12 + 6000·(295 − 245.7)² = 1.504 × 10⁷ mm⁴

web: 10·280³/12 + 2800·(140 − 245.7)² = 4.957 × 10⁷ mm⁴

I_x = 1.504 × 10⁷ + 4.957 × 10⁷ = 6.461 × 10⁷ mm⁴ = 64.61 × 10⁶ mm⁴

Answer

I_x = 64.61 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 5.634 kN·m

Centroid

ȳ = (6600·115 + 1000·50)/7600 = 106.4 mm

top edge: c = 130 − 106.4 = 23.55 mm

Second moment of area

I = 220·30³/12 + 6600·8.553² + 10·100³/12 + 1000·56.45² = 4.997 × 10⁶ mm⁴

Moment allowed by each edge

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

top edge: M = 30·4.997 × 10⁶/23.55 = 6.365 kN·m

bottom edge: M = 120·4.997 × 10⁶/106.4 = 5.634 kN·m

Governing value

M_allow = min = 5.634 kN·m (bottom edge)

Answer

M_allow = 5.634 kN·m

9.Mohr’s circle & stress transformation

Answer: θ_p = 7.018 °

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(2·20, 80 − −80) = 14.04°

Angle on the element

θ_p = 7.018°

Answer

θ_p = 7.018°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 23.66 MPa

Transformation equation

σ_θ = 10 + 10·0.5 + 10·0.866 = 23.66 MPa

Answer

σ_θ = 23.66 MPa

11.Beam deflection

Answer: δ_max = 19.85 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 = 2 N/mm, L = 6000 mm

EI = 1.7 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·2·6000⁴/(384·1.7 × 10¹²) = 19.85 mm

Answer

δ_max = 19.85 mm

12.Beam deflection

Answer: h_req = 234.8 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.

w = 4 N/mm, L = 4500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/250 = 4500/250 = 18 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·4·4500⁴/(384·11000·18) = 1.079 × 10⁸ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.079 × 10⁸/100)^1/3 = 234.8 mm

Answer

h_req = 234.8 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 10.53 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 = 50·20³/12 = 33333 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·33333/2500² = 10528 N = 10.53 kN

Answer

P_cr = 10.53 kN

14.Column buckling

Answer: d = 67.61 mm

Required critical load

P_cr = n·P = 2.5·90000 = 225000 N

KL = 2·1500 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·225000·3000²/(π³·200000))^(1/4) = 67.61 mm

Answer

d = 67.61 mm

Check: the slenderness KL/r is 177.5, so the column is slender and Euler applies.

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