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 steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 GPa). Both are 400 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 100 GPaL = 400 mm

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 200 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 200 mm, E = 100 GPaP = 50 kN

σ₂ = MPa

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 250 mm rotate 0.72° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 20 N·md = 15 mmL = 250 mmφ = 0.72°

G = GPa

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 3000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 3000 rpmd_o = 75 mmd_i = 53 mm

τ_max = MPa

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.

40 kN @ 2 m6 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 w over its whole span and a point load P. Find the maximum bending moment M_max in the beam.

L = 10 mP = 50 kN at 4 mw = 12 kN/m from 0 to 10 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 120 mm × 300 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 195 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 300 mmd = 30 mmhole centre: y = 195 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 10 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 260 mmt_f = 25 mmt_w = 10 mmh_w = 220 mmM = 10 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 30 MPa, σᵧ = 60 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 30 MPaσ_y = 60 MPaτ_xy = −30 MPa

θ_s = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 10 MPa, σᵧ = 80 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 45° clockwise from the x-axis.

σ_x = 10 MPaσ_y = 80 MPaτ_xy = 40 MPaθ = -45°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 1.5 m from A. Find the deflection δ_P under the load.

P = 30 kNa = 1.5 mL = 3 mE = 11 GPa (timber)I = 57 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a point load P at midspan. EI is constant. Find the size of the bending moment M_A at the fixed end.

L = 8 mP = 55 kN at 4 m

M_A = kN·m

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 = 1.5 mE = 200 GPa (steel)b × h = 30 × 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 5 mE = 70 GPa (aluminium)d = 80 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.03587 mm

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·40²/4 = 1257 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 100·10³·3063 = 5.576 × 10⁸ N

Shortening

δ = 50000·400/5.576 × 10⁸ = 0.03587 mm

Answer

δ = 0.03587 mm

2.Axial stress & strain

Answer: σ₂ = 41.75 MPa

Stiffnesses

k₁ = 200000·490.9/400 = 245437 N/mm

k₂ = 100000·706.9/200 = 353429 N/mm

Share of segment 2

F₂ = 50000·353429/598866 = 29508 N

Stress

σ₂ = 29508/706.9 = 41.75 MPa (compression)

Answer

σ₂ = 41.75 MPa

3.Torsion of shafts

Answer: G = 80.06 GPa

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 moment and angle

J = π·15⁴/32 = 4970 mm⁴

φ = 0.72·π/180 = 0.01257 rad

Shear modulus

T = 20 N·m = 20000 N·mm

G = TL/(Jφ) = 20000·250/(4970·0.01257) = 80056 MPa

G = 80.06 GPa

Answer

G = 80.06 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: τ_max = 3.84 MPa

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 75000/314.2 = 238.7 N·m

Polar second moment of area

J = π(75⁴ − 53⁴)/32 = 2.332 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 238732·37.5/2.332 × 10⁶ = 3.84 MPa

Answer

τ_max = 3.84 MPa

5.Shear force & bending moment

Answer: C (x = 2 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 = 44.67 kN, R₂ = 31.33 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 = 41.67, M_C = 77.33 kN·m, M_D = 67, M_E = 50.67, M_F = 28.33, M_G = 0

Answer

Mark C: M_C = 77.33 kN·m

6.Shear force & bending moment

Answer: M_max = 264 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.

UDL resultant: W = w·ℓ = 12·10 = 120 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 50·4 + 12·10·5

R_B = 800/10 = 80 kN

ΣF_y = 0: R_A = 170 − 80 = 90 kN

Where is V = 0?

V(4⁻) = R_A − w·4 = 42 kN

V(4⁺) = 42 − 50 = −8 kN

The shear changes sign at the load, so the maximum is under the load.

Maximum moment

M_max = R_A·4 − w·4²/2 = 90·4 − 12·4²/2 = 264 kN·m

Answer

M_max = 264 kN·m

It occurs at x = 4 m from A.

7.Centroid & second moment of area

Answer: I_x = 268.5 × 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 = (36000·150 − 706.9·195)/(36000 − 706.9) = 149.1 mm

Full plate about ȳ

120·300³/12 + 36000·0.9013² = 270 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·45.9² = 1.529 × 10⁶ mm⁴

Subtract

I_x = 268.5 × 10⁶ mm⁴

Answer

I_x = 268.5 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 59.49 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 = 260·25 = 6500 mm², y_f = 220 + 25/2 = 232.5 mm

web: A_w = 10·220 = 2200 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (6500·232.5 + 2200·110)/8700 = 201.5 mm

Parallel-axis theorem

flange: 260·25³/12 + 6500·(232.5 − 201.5)² = 6.576 × 10⁶ mm⁴

web: 10·220³/12 + 2200·(110 − 201.5)² = 2.73 × 10⁷ mm⁴

I_x = 6.576 × 10⁶ + 2.73 × 10⁷ = 3.388 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 201.5 mm

σ = M·y/I = 1 × 10⁷·201.5/3.388 × 10⁷ = 59.49 MPa

Answer

σ_bottom = 59.49 MPa

9.Mohr’s circle & stress transformation

Answer: θ_s = 76.72 °

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

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −30/33.54 = −0.8944, sin 2θ_s = −(−15)/33.54 = 0.4472

2θ_s = atan2(15, −30) = 153.4°

Angle on the element

θ_s = 76.72°

check: θ_s − θ_p = 76.72 − (−58.28) = 135°

Answer

θ_s = 76.72°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 5 MPa

Transformation equation

σ_θ = 45 + −35·0 + 40·−1 = 5 MPa

Answer

σ_θ = 5 MPa

11.Beam deflection

Answer: δ_P = 53.83 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 = 30000 N, a = 1500 mm, L = 3000 mm

EI = 6.27 × 10¹¹ N·mm²

Deflection under the load

δ_P = Pa³/(3EI) = 30000·1500³/(3·6.27 × 10¹¹) = 53.83 mm

Answer

δ_P = 53.83 mm

12.Beam deflection

Answer: M_A = 82.5 kN·m

Remove the prop

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

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

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) = 5PL³/(48EI) → R_B = 5P/16 = 5·55/16 = 17.19 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = P·L/2 − R_B·L = 220 − 137.5 = 82.5 kN·m

Answer

M_A = 82.5 kN·m

13.Column buckling

Answer: P_cr = 157.9 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·180000/1500² = 157914 N = 157.9 kN

Answer

P_cr = 157.9 kN

14.Column buckling

Answer: σ_cr = 44.22 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 0.5·5000 = 2500 mm

KL/r = 2500/20 = 125

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/125² = 44.22 MPa

Answer

σ_cr = 44.22 MPa

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