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 = 50 kNd = 10 mmL = 3 mE = 70 GPa

ε = µε

2.Axial stress & strainHard

A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of bronze (diameter 40 mm, length 400 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 400 mm, E = 110 GPaP = 50 kN

σ₂ = 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 = 750 N·md = 40 mmL = 3 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 300 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 30 kWN = 300 rpmτ_allow = 40 MPa

d = mm

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 50 kN·m at x = 1.5 m from A, and no other load. Find the bending moment M_x just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.

L = 4 mM₀ = 50 kN·m clockwise, at 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.

40 kN @ 6 m50 kN @ 2 m12 kN/m, 0–4 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 = 280 mmt_f = 15 mmt_w = 14 mmh_w = 300 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 10 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 120 mmt_f = 10 mmt_w = 10 mmh_w = 300 mmM = 10 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 130 MPa, σᵧ = −10 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 40° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 130 MPaσ_y = −10 MPaτ_xy = −10 MPaθ = 40°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −50 MPa, σᵧ = 70 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.

σ_x = −50 MPaσ_y = 70 MPaτ_xy = 20 MPaθ = 60°

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 = 10 kN/mL = 8 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1.5 m from support B. Find the maximum deflection δ_max of the beam.

P = 24 kNb = 1.5 mL = 4 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

δ_max = mm

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 = 4 mE = 200 GPa (steel)d = 50 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 6 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: ε = 9095 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·10²/4 = 78.54 mm²

σ = P/A = 50·1000/78.54 = 636.6 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 636.6/70000 = 0.009095 = 9095 µε

Answer

ε = 9095 µε

2.Axial stress & strain

Answer: σ₂ = 23.27 MPa

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.

Stiffnesses

k₁ = 200000·490.9/400 = 245437 N/mm

k₂ = 110000·1257/400 = 345575 N/mm

Share of segment 2

F₂ = 50000·345575/591012 = 29236 N

Stress

σ₂ = 29236/1257 = 23.27 MPa (compression)

Answer

σ₂ = 23.27 MPa

3.Torsion of shafts

Answer: φ = 19.73 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·3000/(26000·251327) = 0.3443 rad

φ = 0.3443·180/π = 19.73°

Answer

φ = 19.73°

4.Torsion of shafts

Answer: d = 49.54 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 30000/31.42 = 954.9 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·954930/(π·40)) = 49.54 mm

Answer

d = 49.54 mm

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

5.Shear force & bending moment

Answer: M_x = 31.25 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. An applied couple makes the bending moment diagram jump by the size of the couple.

Support reactions

Moments about A, counter-clockwise positive: R_B·L − M₀ = 0. Vertical balance: R_A + R_B = 0.

R_B = M₀/L = 50/4 = 12.5 kN (upward)

R_A = −R_B = −12.5 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −12.5·1.5 + 50 = 31.25 kN·m

Answer

M_x = 31.25 kN·m

6.Shear force & bending moment

Answer: E (x = 4 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 kN, R₂ = 109 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 = 23, M_C = 34, M_D = −17, M_E = −80 kN·m, M_F = −40, M_G = 0

Answer

Mark E: M_E = −80 kN·m

7.Centroid & second moment of area

Answer: I_x = 83.67 × 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 = 280·15 = 4200 mm², y_f = 300 + 15/2 = 307.5 mm

web: A_w = 14·300 = 4200 mm², y_w = 300/2 = 150 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4200·307.5 + 4200·150)/8400 = 228.8 mm

Parallel-axis theorem

flange: 280·15³/12 + 4200·(307.5 − 228.8)² = 2.613 × 10⁷ mm⁴

web: 14·300³/12 + 4200·(150 − 228.8)² = 5.755 × 10⁷ mm⁴

I_x = 2.613 × 10⁷ + 5.755 × 10⁷ = 8.367 × 10⁷ mm⁴ = 83.67 × 10⁶ mm⁴

Answer

I_x = 83.67 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 26.85 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 120·10 = 1200 mm², y_f = 300 + 10/2 = 305 mm

web: A_w = 10·300 = 3000 mm², y_w = 300/2 = 150 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (1200·305 + 3000·150)/4200 = 194.3 mm

Parallel-axis theorem

flange: 120·10³/12 + 1200·(305 − 194.3)² = 1.472 × 10⁷ mm⁴

web: 10·300³/12 + 3000·(150 − 194.3)² = 2.838 × 10⁷ mm⁴

I_x = 1.472 × 10⁷ + 2.838 × 10⁷ = 4.31 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 310 − 194.3 = 115.7 mm

σ = M·y/I = 1 × 10⁷·115.7/4.31 × 10⁷ = 26.85 MPa

Answer

σ_top = 26.85 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = 57.69 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.

Transformation equation for the y′ face

σ_y′ = 60 − 70·0.1736 − (−10)·0.9848 = 57.69 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 62.31 MPa, σ_x′ + σ_y′ = 120 = σ_x + σ_y

Answer

σ_y′ = 57.69 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 57.32 MPa

Transformation equation

σ_θ = 10 + −60·−0.5 + 20·0.866 = 57.32 MPa

Answer

σ_θ = 57.32 MPa

11.Beam deflection

Answer: δ_max = 63.49 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 = 10 N/mm, L = 8000 mm

EI = 8.4 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·10·8000⁴/(384·8.4 × 10¹²) = 63.49 mm

Answer

δ_max = 63.49 mm

12.Beam deflection

Answer: δ_max = 6.399 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 = 24000 N, b = 1500 mm, L = 4000 mm

EI = 4.6 × 10¹² N·mm²

Position of the maximum

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

x = √((L² − b²)/3) = √((4000² − 1500²)/3) = 2141 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 6.399 mm

Answer

δ_max = 6.399 mm

13.Column buckling

Answer: P_cr = 9.462 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/8000² = 9462 N = 9.462 kN

Answer

P_cr = 9.462 kN

14.Column buckling

Answer: σ_cr = 5.483 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 = 1·6000 = 6000 mm

KL/r = 6000/10 = 600

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/600² = 5.483 MPa

Answer

σ_cr = 5.483 MPa

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