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 copper bar under the axial load.

P = 15 kNd = 20 mmL = 2 mE = 117 GPa

δ = mm

2.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø100/Ø70 mm, E = 70 GPa, σ_allow = 70 MPa

P_max = kN

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 60 mm and length 2.5 m; segment BC has diameter 50 mm and length 2.5 m. A torque of 200 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 200 N·md₁ = 60 mm, L₁ = 2.5 md₂ = 50 mm, L₂ = 2.5 mG = 39 GPa

φ_C = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 40 mm is fixed at A. A torque of 200 N·m acts at B, 1.2 m from A, and a torque of 500 N·m in the same direction acts at the free end C, a further 2 m along. Find the maximum shear stress τ_max in the shaft.

d = 40 mmT_B = 200 N·mT_C = 500 N·mAB = 1.2 mBC = 2 m

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.

35 kN @ 6 m6 kN/m, 0–8 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)

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.

25 kN @ 0 m25 kN @ 12 m35 kN @ 4 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

Find the second moment of area I_x of the T-section about the horizontal axis x through its centroid.

b_f = 180 mmt_f = 10 mmt_w = 18 mmh_w = 180 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 40 MPa in tension and 150 MPa in compression. Find the largest hogging bending moment M_allow the beam can carry (give its size).

b_f = 300 mmt_f = 30 mmt_w = 8 mmh_w = 300 mmσ_t,allow = 40 MPaσ_c,allow = 150 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −50 MPa, σᵧ = 100 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 = −50 MPaσ_y = 100 MPaτ_xy = −20 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 60 mm carries an axial tensile force of 40 kN together with a torque of 500 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 60 mmF = 40 kNT = 500 N·m

σ₁ = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 2 m from support A. Find the deflection δ_P under the load.

P = 16 kNa = 2 mL = 8 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 10 kN/mL = 6 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 1 mE = 200 GPa (steel)b × h = 25 × 100 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 15 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 15 kNn = 2L = 2 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.8162 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 = 15·1000/314.2 = 47.75 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 47.75/117000 = 0.0004081 = 408.1 µε

Elongation

δ = ε·L = 0.0004081·2000 mm = 0.8162 mm

Answer

δ = 0.8162 mm

2.Axial stress & strain

Answer: P_max = 531.7 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 + 70·10³·4006 = 5.317 × 10⁸ N

Governing strain

ε = min(0.00125, 0.001) = 0.001 (aluminium)

Largest load

P_max = 0.001·5.317 × 10⁸ = 531715 N = 531.7 kN

Answer

P_max = 531.7 kN

3.Torsion of shafts

Answer: φ_C = 1.774 °

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 moments

J₁ = π·60⁴/32 = 1.272 × 10⁶ mm⁴

J₂ = π·50⁴/32 = 613592 mm⁴

Twist of each segment

T = 200 N·m = 200000 N·mm

G = 39 GPa = 39000 MPa

φ_AB = 200000·2500/(39000·1.272 × 10⁶) = 0.01008 rad

φ_BC = 200000·2500/(39000·613592) = 0.02089 rad

Total twist

φ_C = 0.01008 + 0.02089 = 0.03097 rad = 1.774°

Answer

φ_C = 1.774°

4.Torsion of shafts

Answer: τ_max = 55.7 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 500 N·m

T_AB = T_C + T_B = 500 + 200 = 700 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 700 N·m

τ_max = 16·700000/(π·40³) = 55.7 MPa

Answer

τ_max = 55.7 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 = 49.5 kN, R₂ = 33.5 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 = 87, M_C = 150, M_D = 189 kN·m, M_E = 134, M_F = 67, M_G = 0

Answer

Mark D: M_D = 189 kN·m

6.Shear force & bending moment

Answer: E (x = 8 m)

Support reactions

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

R₁ = 40 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 = −50, M_C = −20, M_D = −60, M_E = −100 kN·m, M_F = −50, M_G = 0

Answer

Mark E: M_E = −100 kN·m

7.Centroid & second moment of area

Answer: I_x = 19.21 × 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 = 180·10 = 1800 mm², y_f = 180 + 10/2 = 185 mm

web: A_w = 18·180 = 3240 mm², y_w = 180/2 = 90 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1800·185 + 3240·90)/5040 = 123.9 mm

Parallel-axis theorem

flange: 180·10³/12 + 1800·(185 − 123.9)² = 6.728 × 10⁶ mm⁴

web: 18·180³/12 + 3240·(90 − 123.9)² = 1.248 × 10⁷ mm⁴

I_x = 6.728 × 10⁶ + 1.248 × 10⁷ = 1.921 × 10⁷ mm⁴ = 19.21 × 10⁶ mm⁴

Answer

I_x = 19.21 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 37.6 kN·m

Centroid

ȳ = (9000·315 + 2400·150)/11400 = 280.3 mm

top edge: c = 330 − 280.3 = 49.74 mm

Second moment of area

I = 300·30³/12 + 9000·34.74² + 8·300³/12 + 2400·130.3² = 7.026 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 40·7.026 × 10⁷/49.74 = 56.5 kN·m

bottom edge: M = 150·7.026 × 10⁷/280.3 = 37.6 kN·m

Governing value

M_allow = min = 37.6 kN·m (bottom edge)

Answer

M_allow = 37.6 kN·m

9.Mohr’s circle & stress transformation

Answer: θ_p = −82.53 °

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, −50 − 100) = −165.1°

Angle on the element

θ_p = −82.53°

Answer

θ_p = −82.53°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ₁ = 20.82 MPa

Stresses at the surface

σ = F/A = 40000/(π·60²/4) = 40000/2827 = 14.15 MPa

τ = 16T/(πd³) = 16·500000/(π·60³) = 11.79 MPa

Mohr’s circle

C = σ/2 = 7.074 MPa

R = √(7.074² + 11.79²) = 13.75 MPa

σ₁ = C + R = 20.82 MPa

σ₂ = C − R = −6.675 MPa

Answer

σ₁ = 20.82 MPa

11.Beam deflection

Answer: δ_P = 38.1 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 = 16000 N, a = 2000 mm, b = 6000 mm, L = 8000 mm

EI = 2.52 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 16000·2000²·6000²/(3·2.52 × 10¹²·8000) = 38.1 mm

Answer

δ_P = 38.1 mm

12.Beam deflection

Answer: I_req = 50.63 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 6000/360 = 16.67 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·10·6000⁴/(384·200000·16.67) = 5.063 × 10⁷ mm⁴

I_req = 50.63 × 10⁶ mm⁴

Answer

I_req = 50.63 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 64.26 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 = 100·25³/12 = 130208 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·130208/2000² = 64255 N = 64.26 kN

Answer

P_cr = 64.26 kN

14.Column buckling

Answer: d = 36.29 mm

Required critical load

P_cr = n·P = 2·15000 = 30000 N

KL = 0.7·2000 = 1400 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·30000·1400²/(π³·70000))^(1/4) = 36.29 mm

Answer

d = 36.29 mm

Check: the slenderness KL/r is 154.3, so the column is slender and Euler applies.

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