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 = 120 kNd = 20 mmL = 2 mE = 200 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaP = 30 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid brass 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 = 50 mmL = 1.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 5 kW at 300 rpm. The shear stress must not exceed 50 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.

P = 5 kWN = 300 rpmτ_allow = 50 MPaθ_allow = 0.25°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 1 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 10 kN/m from 0 to 5 mx = 1 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 @ 0 m35 kN @ 12 m30 kN @ 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)

7.Centroid & second moment of areaMedium

A 100 mm × 200 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 60 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 100 mmh = 200 mmd = 30 mmhole centre: y = 60 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 30 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 300 mmt_f = 30 mmt_w = 14 mmh_w = 280 mmM = 30 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 70 MPa, σᵧ = 90 MPa and τₓᵧ = −50 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 = 70 MPaσ_y = 90 MPaτ_xy = −50 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 130 MPa, σᵧ = −40 MPa and τₓᵧ = 50 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = 130 MPaσ_y = −40 MPaτ_xy = 50 MPa

σ_vM = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δ_P under the load.

P = 12 kNa = 1 mL = 5 mE = 70 GPa (aluminium)I = 57 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported beam carries a point load P at b = 1 m from support B. Find the maximum deflection δ_max of the beam.

P = 17 kNb = 1 mL = 5 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 5 mE = 200 GPa (steel)d = 80 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 10 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 10 kNn = 2.5L = 5 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 3.82 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 = 120·1000/314.2 = 382 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 382/200000 = 0.00191 = 1910 µε

Elongation

δ = ε·L = 0.00191·2000 mm = 3.82 mm

Answer

δ = 3.82 mm

2.Axial stress & strain

Answer: σ₂ = 13.26 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 = 200000·1257/500 = 502655 N/mm

k₂ = 100000·1257/200 = 628319 N/mm

Share of segment 2

F₂ = 30000·628319/1.131 × 10⁶ = 16667 N

Stress

σ₂ = 16667/1257 = 13.26 MPa (compression)

Answer

σ₂ = 13.26 MPa

3.Torsion of shafts

Answer: φ = 2.694 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1500/(39000·613592) = 0.04701 rad

φ = 0.04701·180/π = 2.694°

Answer

φ = 2.694°

4.Torsion of shafts

Answer: d = 55.56 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 5000/31.42 = 159.2 N·m

Diameter for strength

T = 159.2 N·m = 159155 N·mm

d_τ = ∛(16·159155/(π·50)) = 25.31 mm

Diameter for stiffness

θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·159155/(π·39000·4.363 × 10⁻⁶))^1/4 = 55.56 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(25.31, 55.56) = 55.56 mm

Answer

d = 55.56 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = 20 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.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·5 = 50 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 10·5·2.5

R_B = 125/5 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 1 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 = 25·1 − 10·1·0.5 = 20 kN·m

Answer

M_x = 20 kN·m

6.Shear force & bending moment

Answer: F (x = 10 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 = 30 kN, R₂ = 60 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 = −40, M_D = −30, M_E = −20, M_F = −70 kN·m, M_G = 0

Answer

Mark F: M_F = −70 kN·m

7.Centroid & second moment of area

Answer: I_x = 65.45 × 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 = (20000·100 − 706.9·60)/(20000 − 706.9) = 101.5 mm

Full plate about ȳ

100·200³/12 + 20000·−1.466² = 66.71 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·−41.47² = 1.255 × 10⁶ mm⁴

Subtract

I_x = 65.45 × 10⁶ mm⁴

Answer

I_x = 65.45 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 80.96 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 = 300·30 = 9000 mm², y_f = 280 + 30/2 = 295 mm

web: A_w = 14·280 = 3920 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (9000·295 + 3920·140)/12920 = 248 mm

Parallel-axis theorem

flange: 300·30³/12 + 9000·(295 − 248)² = 2.058 × 10⁷ mm⁴

web: 14·280³/12 + 3920·(140 − 248)² = 7.131 × 10⁷ mm⁴

I_x = 2.058 × 10⁷ + 7.131 × 10⁷ = 9.189 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 248 mm

σ = M·y/I = 3 × 10⁷·248/9.189 × 10⁷ = 80.96 MPa

Answer

σ_bottom = 80.96 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = −50.65 °

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·−50, 70 − 90) = −101.3°

Angle on the element

θ_p = −50.65°

Answer

θ_p = −50.65°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_vM = 176.6 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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 45 MPa, R = 98.62 MPa

σ₁ = 143.6 MPa, σ₂ = −53.62 MPa, σ₃ = 0

Von Mises

σ_vM = √(143.6² − 143.6·(−53.62) + (−53.62)²) = 176.6 MPa

Answer

σ_vM = 176.6 MPa

11.Beam deflection

Answer: δ_P = 3.208 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 = 12000 N, a = 1000 mm, b = 4000 mm, L = 5000 mm

EI = 3.99 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 12000·1000²·4000²/(3·3.99 × 10¹²·5000) = 3.208 mm

Answer

δ_P = 3.208 mm

12.Beam deflection

Answer: δ_max = 5.575 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 = 17000 N, b = 1000 mm, L = 5000 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) = √((5000² − 1000²)/3) = 2828 mm

Maximum deflection

δ_max = Pb(L² − b²)^3/2/(9√3·EIL) = 5.575 mm

Answer

δ_max = 5.575 mm

13.Column buckling

Answer: P_cr = 158.8 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 1·5000 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/5000² = 158752 N = 158.8 kN

Answer

P_cr = 158.8 kN

14.Column buckling

Answer: d = 54.82 mm

Required critical load

P_cr = n·P = 2.5·10000 = 25000 N

KL = 0.7·5000 = 3500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·25000·3500²/(π³·70000))^(1/4) = 54.82 mm

Answer

d = 54.82 mm

Check: the slenderness KL/r is 255.4, so the column is slender and Euler applies.

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