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 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 500 mm, E = 100 GPaδ = 0.2 mm

P = kN

2.Axial stress & strainHard

A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of brass (diameter 60 mm, length 500 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: brass, Ø60 mm, 500 mm, E = 100 GPaP = 60 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 750 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

T = 750 N·md_i/d_o = 0.6τ_allow = 40 MPa

d_o = mm

4.Torsion of shaftsHard

A solid steel shaft of diameter 40 mm is fixed at A. A torque of 200 N·m acts at B, 2.5 m from A, and a torque of 1500 N·m in the same direction acts at the free end C, a further 1.5 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 40 mmT_B = 200 N·mT_C = 1500 N·mAB = 2.5 mBC = 1.5 mG = 80 GPa

φ_C = °

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 10 kN·m at x = 4 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 = 8 mM₀ = 10 kN·m clockwise, at 4 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.

25 kN @ 6 m50 kN @ 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

A symmetric I-section has flanges 100 mm wide and 14 mm thick, an overall depth of 260 mm and a web 14 mm thick. Find its second moment of area I_x about the horizontal centroidal axis x.

B = 100 mmH = 260 mmt_f = 14 mmt_w = 14 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 50 MPa in tension and 120 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 100 mmt_f = 25 mmt_w = 20 mmh_w = 200 mmσ_t,allow = 50 MPaσ_c,allow = 120 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 50 MPa, σᵧ = 10 MPa and τₓᵧ = −60 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 50 MPaσ_y = 10 MPaτ_xy = −60 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 10 MPa, σᵧ = 100 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σ_x = 10 MPaσ_y = 100 MPaτ_xy = −10 MPaθ = 50°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.

w = 4 kN/mL = 5 mE = 70 GPa (aluminium)I = 120 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

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

P = 2 kNb = 1.5 mL = 4 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ_mid = 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 = 6 mE = 200 GPa (steel)b × h = 40 × 100 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at both ends. It must carry 140 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 140 kNn = 3L = 4 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 19.33 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 = 314.2 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(100000·1257) = 3.979 × 10⁻⁶ mm/N

Load

P = 0.2/1.035 × 10⁻⁵ = 19333 N = 19.33 kN

Answer

P = 19.33 kN

2.Axial stress & strain

Answer: σ₂ = 17.9 MPa

Stiffnesses

k₁ = 200000·314.2/600 = 104720 N/mm

k₂ = 100000·2827/500 = 565487 N/mm

Share of segment 2

F₂ = 60000·565487/670206 = 50625 N

Stress

σ₂ = 50625/2827 = 17.9 MPa (compression)

Answer

σ₂ = 17.9 MPa

3.Torsion of shafts

Answer: d_o = 47.87 mm

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.

Outer diameter from the allowable stress

T = 750 N·m = 750000 N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·750000/(π·40·0.8704)) = 47.87 mm

d_i = 0.6·47.87 = 28.72 mm

Answer

d_o = 47.87 mm

A solid shaft would need d = 45.71 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: φ_C = 18.52 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 1500 N·m

T_AB = T_C + T_B = 1500 + 200 = 1700 N·m

Twist of each segment

J = π·40⁴/32 = 251327 mm⁴

G = 80 GPa = 80000 MPa

φ_AB = 1.7 × 10⁶·2500/(80000·251327) = 0.2114 rad

φ_BC = 1.5 × 10⁶·1500/(80000·251327) = 0.1119 rad

Twist of end C

φ_C = 0.2114 + 0.1119 = 0.3233 rad → |φ_C| = 18.52°

Answer

φ_C = 18.52°

5.Shear force & bending moment

Answer: M_x = 5 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 = 10/8 = 1.25 kN (upward)

R_A = −R_B = −1.25 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −1.25·4 + 10 = 5 kN·m

Answer

M_x = 5 kN·m

6.Shear force & bending moment

Answer: F (x = 5 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 = 5 kN, R₂ = 70 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 = 5, M_C = 10, M_D = 15, M_E = 20, M_F = −25 kN·m, M_G = 0

Answer

Mark F: M_F = −25 kN·m

7.Centroid & second moment of area

Answer: I_x = 56.98 × 10⁶ mm⁴

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.

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 = 260 − 2·14 = 232 mm

I_x = (B·H³ − (B − t_w)·h_w³)/12 = (100·260³ − 86·232³)/12

I_x = 5.698 × 10⁷ mm⁴ = 56.98 × 10⁶ mm⁴

Answer

I_x = 56.98 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 11.49 kN·m

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.

Centroid

ȳ = (2500·212.5 + 4000·100)/6500 = 143.3 mm

top edge: c = 225 − 143.3 = 81.73 mm

Second moment of area

I = 100·25³/12 + 2500·69.23² + 20·200³/12 + 4000·43.27² = 3.293 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 120·3.293 × 10⁷/81.73 = 48.36 kN·m

bottom edge: M = 50·3.293 × 10⁷/143.3 = 11.49 kN·m

Governing value

M_allow = min = 11.49 kN·m (bottom edge)

Answer

M_allow = 11.49 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_max = 63.25 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.

Radius of Mohr’s circle

τ_max = R = √(((50 − 10)/2)² + −60²) = 63.25 MPa

Answer

τ_max = 63.25 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 52.97 MPa

Transformation equation

σ_θ = 55 + −45·−0.1736 + −10·0.9848 = 52.97 MPa

Answer

σ_θ = 52.97 MPa

11.Beam deflection

Answer: δ = 37.2 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 = 4 N/mm, L = 5000 mm

EI = 8.4 × 10¹² N·mm²

Deflection

δ = wL⁴/(8EI) = 4·5000⁴/(8·8.4 × 10¹²) = 37.2 mm

Answer

δ = 37.2 mm

12.Beam deflection

Answer: δ_mid = 1.385 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 = 2000 N, b = 1500 mm, L = 4000 mm

EI = 1.76 × 10¹² N·mm²

Midspan deflection

δ_mid = Pb(3L² − 4b²)/(48EI) = 2000·1500·(3·4000² − 4·1500²)/(48·1.76 × 10¹²) = 1.385 mm

Answer

δ_mid = 1.385 mm

13.Column buckling

Answer: P_cr = 7.311 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 2·6000 = 12000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·533333/12000² = 7311 N = 7.311 kN

Answer

P_cr = 7.311 kN

14.Column buckling

Answer: d = 83.89 mm

Required critical load

P_cr = n·P = 3·140000 = 420000 N

KL = 0.5·4000 = 2000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·420000·2000²/(π³·70000))^(1/4) = 83.89 mm

Answer

d = 83.89 mm

Check: the slenderness KL/r is 95.36, so the column is slender and Euler applies.

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