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 steel bar, in microstrain (µε = 10⁻⁶).

P = 100 kNd = 16 mmL = 2 mE = 200 GPa

ε = µε

2.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 60 mm, E = 70 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø100/Ø60 mm, E = 70 GPa, σ_allow = 80 MPa

P_max = kN

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 80 mm, inner diameter 60 mm, length 1 m) is fixed at one end and twisted by a torque of 2000 N·m at the other. Find the angle of twist φ in degrees.

T = 2000 N·md_o = 80 mmd_i = 60 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft of diameter 30 mm is fixed at A. A torque of 400 N·m acts at B, 3 m from A, and a torque of 250 N·m in the opposite direction acts at the free end C, a further 3 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 30 mmT_B = 400 N·mT_C = 250 N·mAB = 3 mBC = 3 mG = 80 GPa

φ_C = °

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.

30 kN @ 4 m4 kN/m, 0–6 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 over part of its span. Find the distance x₀ from A to the section where the bending moment is largest.

L = 6 mw = 12 kN/m from 2 to 6 m

x₀ = m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 90 mm and inside diameter d = 78 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 90 mmd = 78 mm

r = mm

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 = 100 mmt_f = 20 mmt_w = 10 mmh_w = 140 mmM = 10 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 140 MPa, σᵧ = −50 MPa and τₓᵧ = 30 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 = 140 MPaσ_y = −50 MPaτ_xy = 30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 20 MPa, σᵧ = 50 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.

σ_x = 20 MPaσ_y = 50 MPaτ_xy = −10 MPaθ = 25°

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 = 6 kN/mL = 3 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 6 mw = 5 kN/m

R_B = kN

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 = 6 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 65 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 65 kNn = 2.5L = 2 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 2487 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·16²/4 = 201.1 mm²

σ = P/A = 100·1000/201.1 = 497.4 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 497.4/200000 = 0.002487 = 2487 µε

Answer

ε = 2487 µε

2.Axial stress & strain

Answer: P_max = 473.9 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·20²/4 = 314.2 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 70·10³·5027 = 4.147 × 10⁸ N

Governing strain

ε = min(0.00125, 0.001143) = 0.001143 (aluminium)

Largest load

P_max = 0.001143·4.147 × 10⁸ = 473932 N = 473.9 kN

Answer

P_max = 473.9 kN

3.Torsion of shafts

Answer: φ = 1.069 °

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(80⁴ − 60⁴)/32 = 2.749 × 10⁶ mm⁴

Angle of twist

T = 2000 N·m = 2 × 10⁶ N·mm

L = 1 m = 1000 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 2 × 10⁶·1000/(39000·2.749 × 10⁶) = 0.01866 rad

φ = 0.01866·180/π = 1.069°

Answer

φ = 1.069°

4.Torsion of shafts

Answer: φ_C = 2.702 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 250 N·m

T_AB = T_C − T_B = 250 − 400 = −150 N·m

Twist of each segment

J = π·30⁴/32 = 79522 mm⁴

G = 80 GPa = 80000 MPa

φ_AB = −150000·3000/(80000·79522) = −0.07074 rad

φ_BC = 250000·3000/(80000·79522) = 0.1179 rad

Twist of end C

φ_C = −0.07074 + 0.1179 = 0.04716 rad → |φ_C| = 2.702°

Answer

φ_C = 2.702°

5.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 = 22 kN, R₂ = 32 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 = 20, M_C = 36, M_D = 48, M_E = 56 kN·m, M_F = 30, M_G = 0

Answer

Mark E: M_E = 56 kN·m

6.Shear force & bending moment

Answer: x₀ = 3.333 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 = 12·4 = 48 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·6 = 12·4·4

R_B = 192/6 = 32 kN

ΣF_y = 0: R_A = 48 − 32 = 16 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 2) = 0

x₀ = 2 + 16/12 = 3.333 m

Answer

x₀ = 3.333 m

7.Centroid & second moment of area

Answer: r = 29.77 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 1583 mm²

I = π(D⁴ − d⁴)/64 = 1.404 × 10⁶ mm⁴

Radius of gyration

r = √(I/A) = √(1.404 × 10⁶/1583) = 29.77 mm

Answer

r = 29.77 mm

8.Centroid & second moment of area

Answer: σ_top = 56.32 MPa

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 = 100·20 = 2000 mm², y_f = 140 + 20/2 = 150 mm

web: A_w = 10·140 = 1400 mm², y_w = 140/2 = 70 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2000·150 + 1400·70)/3400 = 117.1 mm

Parallel-axis theorem

flange: 100·20³/12 + 2000·(150 − 117.1)² = 2.237 × 10⁶ mm⁴

web: 10·140³/12 + 1400·(70 − 117.1)² = 5.387 × 10⁶ mm⁴

I_x = 2.237 × 10⁶ + 5.387 × 10⁶ = 7.624 × 10⁶ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 160 − 117.1 = 42.94 mm

σ = M·y/I = 1 × 10⁷·42.94/7.624 × 10⁶ = 56.32 MPa

Answer

σ_top = 56.32 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 8.763 °

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(2·30, 140 − −50) = 17.53°

Angle on the element

θ_p = 8.763°

Answer

θ_p = 8.763°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 17.7 MPa

Transformation equation

σ_θ = 35 + −15·0.6428 + −10·0.766 = 17.7 MPa

Answer

σ_θ = 17.7 MPa

11.Beam deflection

Answer: δ_max = 3.596 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 = 6 N/mm, L = 3000 mm

EI = 1.76 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·6·3000⁴/(384·1.76 × 10¹²) = 3.596 mm

Answer

δ_max = 3.596 mm

12.Beam deflection

Answer: R_B = 11.25 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·5·6/8 = 11.25 kN

Answer

R_B = 11.25 kN

13.Column buckling

Answer: P_cr = 2.412 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 = 1·6000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/6000² = 2412 N = 2.412 kN

Answer

P_cr = 2.412 kN

14.Column buckling

Answer: d = 50.89 mm

Required critical load

P_cr = n·P = 2.5·65000 = 162500 N

KL = 1·2000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·162500·2000²/(π³·200000))^(1/4) = 50.89 mm

Answer

d = 50.89 mm

Check: the slenderness KL/r is 157.2, so the column is slender and Euler applies.

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