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 30 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 400 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø30 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 400 mm, E = 70 GPaδ = 0.3 mm

P = kN

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 400 mm, E = 70 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, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 400 mm, E = 70 GPaP = 80 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A stepped solid brass shaft is fixed at A. Segment AB has diameter 60 mm and length 2.5 m; segment BC has diameter 40 mm and length 2 m. A torque of 250 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 250 N·md₁ = 60 mm, L₁ = 2.5 md₂ = 40 mm, L₂ = 2 mG = 39 GPa

φ_C = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 100 kWN = 500 rpmτ_allow = 50 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 0.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 4 mw = 4 kN/m from 0 to 4 mx = 0.5 m

M_x = kN·m

6.Shear force & bending momentHard

The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the maximum sagging bending moment M_max between the supports.

L = 6 mw = 6 kN/m from 0 to 6 msupports at x = 0 and x = 4 m

M_max = kN·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 = 100 mmt_f = 30 mmt_w = 8 mmh_w = 300 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 200 mm × 150 mm plate has a rectangular cut-out 50 mm wide and 50 mm high, centred on the vertical centre line with its centre 52.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 200 mmh = 150 mmcut-out: 50 × 50 mmhole centre: y = 52.5 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −30 MPa, σᵧ = 20 MPa and τₓᵧ = −20 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 = −30 MPaσ_y = 20 MPaτ_xy = −20 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 800 mm and wall thickness 10 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 800 mmt = 10 mmp = 1.5 MPaα = 55°

|τ_n| = 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 = 4 kN/mL = 4 mE = 11 GPa (timber)I = 36 × 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 = 11 kNw = 15 kN/mL = 2.5 mE = 200 GPa (steel)I = 36 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 2 mE = 200 GPa (steel)b × h = 30 × 60 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 115 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 115 kNn = 3L = 3 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 24.33 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 = 706.9 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(70000·706.9) = 8.084 × 10⁻⁶ mm/N

Load

P = 0.3/1.233 × 10⁻⁵ = 24334 N = 24.33 kN

Answer

P = 24.33 kN

2.Axial stress & strain

Answer: σ₂ = 25.8 MPa

Stiffnesses

k₁ = 200000·1257/600 = 418879 N/mm

k₂ = 70000·706.9/400 = 123700 N/mm

Share of segment 2

F₂ = 80000·123700/542579 = 18239 N

Stress

σ₂ = 18239/706.9 = 25.8 MPa (compression)

Answer

σ₂ = 25.8 MPa

3.Torsion of shafts

Answer: φ_C = 3.644 °

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 moments

J₁ = π·60⁴/32 = 1.272 × 10⁶ mm⁴

J₂ = π·40⁴/32 = 251327 mm⁴

Twist of each segment

T = 250 N·m = 250000 N·mm

G = 39 GPa = 39000 MPa

φ_AB = 250000·2500/(39000·1.272 × 10⁶) = 0.0126 rad

φ_BC = 250000·2000/(39000·251327) = 0.05101 rad

Total twist

φ_C = 0.0126 + 0.05101 = 0.06361 rad = 3.644°

Answer

φ_C = 3.644°

4.Torsion of shafts

Answer: d = 57.94 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 100000/52.36 = 1910 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·1.91 × 10⁶/(π·50)) = 57.94 mm

Answer

d = 57.94 mm

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

5.Shear force & bending moment

Answer: M_x = 3.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 4·4 = 16 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 4·4·2

R_B = 32/4 = 8 kN

ΣF_y = 0: R_A = 16 − 8 = 8 kN

Cut at x = 0.5 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 = 8·0.5 − 4·0.5·0.25 = 3.5 kN·m

Answer

M_x = 3.5 kN·m

6.Shear force & bending moment

Answer: M_max = 6.75 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. 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, including the load on the overhang.

UDL resultant: W = w·ℓ = 6·6 = 36 kN, acting at x̄ = 3 m

ΣM_A = 0: R_B·4 = 6·6·3

R_B = 108/4 = 27 kN

ΣF_y = 0: R_A = 36 − 27 = 9 kN

Zero shear

R_A − w·x₀ = 0 → x₀ = 9/6 = 1.5 m

Maximum sagging moment

M_max = R_A·x₀ − w·x₀²/2 = R_A²/(2w) = 9²/(2·6) = 6.75 kN·m

Answer

M_max = 6.75 kN·m

7.Centroid & second moment of area

Answer: I_x = 54.52 × 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 = 100·30 = 3000 mm², y_f = 300 + 30/2 = 315 mm

web: A_w = 8·300 = 2400 mm², y_w = 300/2 = 150 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3000·315 + 2400·150)/5400 = 241.7 mm

Parallel-axis theorem

flange: 100·30³/12 + 3000·(315 − 241.7)² = 1.636 × 10⁷ mm⁴

web: 8·300³/12 + 2400·(150 − 241.7)² = 3.817 × 10⁷ mm⁴

I_x = 1.636 × 10⁷ + 3.817 × 10⁷ = 5.452 × 10⁷ mm⁴ = 54.52 × 10⁶ mm⁴

Answer

I_x = 54.52 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 54.35 × 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 = (30000·75 − 2500·52.5)/(30000 − 2500) = 77.05 mm

Full plate about ȳ

200·150³/12 + 30000·−2.045² = 56.38 × 10⁶ mm⁴

Hole about ȳ

50·50³/12 + 2500·−24.55² = 2.027 × 10⁶ mm⁴

Subtract

I_x = 54.35 × 10⁶ mm⁴

Answer

I_x = 54.35 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = −70.67 °

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·−20, −30 − 20) = −141.3°

Angle on the element

θ_p = −70.67°

Answer

θ_p = −70.67°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 14.1 MPa

Wall stresses

r = 800/2 = 400 mm

σ_h = pr/t = 1.5·400/10 = 60 MPa

σ_a = pr/(2t) = 30 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

|τ_n| = (60 − 30)·sin 55°·cos 55° = 14.1 MPa

Answer

|τ_n| = 14.1 MPa

11.Beam deflection

Answer: δ_max = 33.67 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 = 4000 mm

EI = 3.96 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·4·4000⁴/(384·3.96 × 10¹¹) = 33.67 mm

Answer

δ_max = 33.67 mm

12.Beam deflection

Answer: δ_max = 1.557 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 = 11000 N, w = 15 N/mm, L = 2500 mm, EI = 7.2 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 11000·2500³/(48·7.2 × 10¹²) = 0.4973 mm

δ_w = 5wL⁴/(384EI) = 5·15·2500⁴/(384·7.2 × 10¹²) = 1.06 mm

Add them

δ_max = 0.4973 + 1.06 = 1.557 mm

Answer

δ_max = 1.557 mm

13.Column buckling

Answer: P_cr = 136 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·135000/1400² = 135959 N = 136 kN

Answer

P_cr = 136 kN

14.Column buckling

Answer: d = 75.24 mm

Required critical load

P_cr = n·P = 3·115000 = 345000 N

KL = 1·3000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·345000·3000²/(π³·200000))^(1/4) = 75.24 mm

Answer

d = 75.24 mm

Check: the slenderness KL/r is 159.5, so the column is slender and Euler applies.

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