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 steel bar under the axial load.

P = 50 kNd = 10 mmL = 0.5 mE = 200 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) and a segment made of copper (diameter 60 mm, length 200 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: copper, Ø60 mm, 200 mm, E = 120 GPaP = 80 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 2000 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter d_o.

T = 2000 N·md_i/d_o = 0.7τ_allow = 60 MPa

d_o = mm

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 750 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 750 rpmd_o = 75 mmd_i = 38 mm

τ_max = MPa

5.Shear force & bending momentMedium

The simply supported beam carries a clockwise couple M₀ = 45 kN·m at x = 2 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 = 5 mM₀ = 45 kN·m clockwise, at 2 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 6.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 15 kN at 4 mw = 8 kN/m from 0 to 10 mx = 6.5 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A 200 mm × 240 mm plate has a circular hole of diameter 60 mm on its vertical centre line, with its centre 72 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 240 mmd = 60 mmhole centre: y = 72 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 5 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 200 mmt_f = 30 mmt_w = 16 mmh_w = 240 mmM = 5 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 110 MPa, σᵧ = 90 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 110 MPaσ_y = 90 MPaτ_xy = −10 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 120 MPa, σᵧ = 60 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.

σ_x = 120 MPaσ_y = 60 MPaτ_xy = 30 MPaθ = 35°

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 = 5 mE = 11 GPa (timber)I = 120 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 9 kNw = 4 kN/mL = 2 mE = 11 GPa (timber)I = 160 × 10⁶ mm⁴

δ_max = mm

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 = 2 mE = 70 GPa (aluminium)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 = 2.5 mE = 200 GPa (steel)d = 60 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 1.592 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·10²/4 = 78.54 mm²

σ = P/A = 50·1000/78.54 = 636.6 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 636.6/200000 = 0.003183 = 3183 µε

Elongation

δ = ε·L = 0.003183·500 mm = 1.592 mm

Answer

δ = 1.592 mm

2.Axial stress & strain

Answer: σ₂ = 26.34 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 = 200000·314.2/500 = 125664 N/mm

k₂ = 120000·2827/200 = 1.696 × 10⁶ N/mm

Share of segment 2

F₂ = 80000·1.696 × 10⁶/1.822 × 10⁶ = 74483 N

Stress

σ₂ = 74483/2827 = 26.34 MPa (compression)

Answer

σ₂ = 26.34 MPa

3.Torsion of shafts

Answer: d_o = 60.68 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 = 2000 N·m = 2 × 10⁶ N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·2 × 10⁶/(π·60·0.7599)) = 60.68 mm

d_i = 0.7·60.68 = 42.47 mm

Answer

d_o = 60.68 mm

A solid shaft would need d = 55.37 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: τ_max = 16.46 MPa

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 100000/78.54 = 1273 N·m

Polar second moment of area

J = π(75⁴ − 38⁴)/32 = 2.902 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 1.273 × 10⁶·37.5/2.902 × 10⁶ = 16.46 MPa

Answer

τ_max = 16.46 MPa

5.Shear force & bending moment

Answer: M_x = 27 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 = 45/5 = 9 kN (upward)

R_A = −R_B = −9 kN (downward)

Cut just right of the couple

The left part now contains R_A and the couple.

M_x = R_A·a + M₀ = −9·2 + 45 = 27 kN·m

Answer

M_x = 27 kN·m

6.Shear force & bending moment

Answer: M_x = 112 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 8·10 = 80 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 15·4 + 8·10·5

R_B = 460/10 = 46 kN

ΣF_y = 0: R_A = 95 − 46 = 49 kN

Cut at x = 6.5 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 49·6.5 − 15·(6.5 − 4) − 8·6.5·3.25 = 112 kN·m

Answer

M_x = 112 kN·m

7.Centroid & second moment of area

Answer: I_x = 222.8 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (48000·120 − 2827·72)/(48000 − 2827) = 123 mm

Full plate about ȳ

200·240³/12 + 48000·−3.004² = 230.8 × 10⁶ mm⁴

Hole about ȳ

π·60⁴/64 + 2827·−51² = 7.992 × 10⁶ mm⁴

Subtract

I_x = 222.8 × 10⁶ mm⁴

Answer

I_x = 222.8 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 16.43 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 = 200·30 = 6000 mm², y_f = 240 + 30/2 = 255 mm

web: A_w = 16·240 = 3840 mm², y_w = 240/2 = 120 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6000·255 + 3840·120)/9840 = 202.3 mm

Parallel-axis theorem

flange: 200·30³/12 + 6000·(255 − 202.3)² = 1.71 × 10⁷ mm⁴

web: 16·240³/12 + 3840·(120 − 202.3)² = 4.445 × 10⁷ mm⁴

I_x = 1.71 × 10⁷ + 4.445 × 10⁷ = 6.156 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 202.3 mm

σ = M·y/I = 5 × 10⁶·202.3/6.156 × 10⁷ = 16.43 MPa

Answer

σ_bottom = 16.43 MPa

9.Mohr’s circle & stress transformation

Answer: τ_max = 14.14 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 = √(((110 − 90)/2)² + −10²) = 14.14 MPa

Answer

τ_max = 14.14 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 128.5 MPa

Transformation equation

σ_θ = 90 + 30·0.342 + 30·0.9397 = 128.5 MPa

Answer

σ_θ = 128.5 MPa

11.Beam deflection

Answer: δ_max = 36.99 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 = 5000 mm

EI = 1.32 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·6·5000⁴/(384·1.32 × 10¹²) = 36.99 mm

Answer

δ_max = 36.99 mm

12.Beam deflection

Answer: δ_max = 1.326 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 = 9000 N, w = 4 N/mm, L = 2000 mm, EI = 1.76 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 9000·2000³/(48·1.76 × 10¹²) = 0.8523 mm

δ_w = 5wL⁴/(384EI) = 5·4·2000⁴/(384·1.76 × 10¹²) = 0.4735 mm

Add them

δ_max = 0.8523 + 0.4735 = 1.326 mm

Answer

δ_max = 1.326 mm

13.Column buckling

Answer: P_cr = 52.99 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 = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·306796/2000² = 52989 N = 52.99 kN

Answer

P_cr = 52.99 kN

14.Column buckling

Answer: σ_cr = 71.06 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 1·2500 = 2500 mm

KL/r = 2500/15 = 166.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/166.7² = 71.06 MPa

Answer

σ_cr = 71.06 MPa

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