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 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 400 mm, E = 100 GPaδ = 0.3 mm

P = kN

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 80 kNsegment 1: Ø30 mm, 1 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, aluminium (E = 70 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid steel 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 = 250 N·md = 40 mmL = 0.8 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 1450 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1450 rpmτ_allow = 60 MPa

d = mm

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.

15 kN @ 3 m6 kN/m, 2–6 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 beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the bending moment M_B at support B. Use sagging moments positive and upward shear on the left face positive.

L = 6.5 mw = 8 kN/m from 0 to 6.5 msupports at x = 0 and x = 5 m

M_B = 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 = 180 mmt_f = 15 mmt_w = 8 mmh_w = 280 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 40 MPa in tension and 90 MPa in compression. Find the largest sagging bending moment M_allow the beam can carry.

b_f = 260 mmt_f = 15 mmt_w = 8 mmh_w = 180 mmσ_t,allow = 40 MPaσ_c,allow = 90 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 100 MPa, σᵧ = −60 MPa and τₓᵧ = −30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 100 MPaσ_y = −60 MPaτ_xy = −30 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A solid shaft of diameter 20 mm carries an axial compressive force of 150 kN together with a torque of 250 N·m. At the surface of the shaft, find the largest principal stress σ₁.

d = 20 mmF = −150 kNT = 250 N·m

σ₁ = MPa

11.Beam deflectionMedium

A cantilever of length L is fixed at A and carries a point load P at a distance a = 0.5 m from A. Find the deflection δ_B at the free end B.

P = 15 kNa = 0.5 mL = 3 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_B = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area I_req that meets this limit.

w = 10 kN/mL = 6 mE = 11 GPa (timber)

I_req = × 10⁶ mm⁴

13.Column bucklingMedium

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

L = 1.5 mE = 70 GPa (aluminium)d = 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: P = 62.83 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 = 1257 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(100000·1257) = 3.183 × 10⁻⁶ mm/N

Load

P = 0.3/4.775 × 10⁻⁶ = 62832 N = 62.83 kN

Answer

P = 62.83 kN

2.Axial stress & strain

Answer: σ_max = 707.4 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·30²/4 = 706.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 113.2 MPa

σ₂ = 707.4 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 707.4 MPa

Answer

σ_max = 707.4 MPa

3.Torsion of shafts

Answer: φ = 0.5699 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·800/(80000·251327) = 0.009947 rad

φ = 0.009947·180/π = 0.5699°

Answer

φ = 0.5699°

4.Torsion of shafts

Answer: d = 20.32 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 15000/151.8 = 98.79 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·98786/(π·60)) = 20.32 mm

Answer

d = 20.32 mm

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

5.Shear force & bending moment

Answer: D (x = 3 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 = 15.5 kN, R₂ = 23.5 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 = 15.5, M_C = 31, M_D = 43.5 kN·m, M_E = 35, M_F = 20.5, M_G = 0

Answer

Mark D: M_D = 43.5 kN·m

6.Shear force & bending moment

Answer: M_B = −9 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 = 8·6.5 = 52 kN, acting at x̄ = 3.25 m

ΣM_A = 0: R_B·5 = 8·6.5·3.25

R_B = 169/5 = 33.8 kN

ΣF_y = 0: R_A = 52 − 33.8 = 18.2 kN

Cut at B and keep the overhang

The load on the overhang, w·a, acts at a/2 from B and makes the beam hog over the support.

M_B = −w·a²/2 = −8·1.5²/2 = −9 kN·m

Answer

M_B = −9 kN·m

7.Centroid & second moment of area

Answer: I_x = 41.32 × 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 = 180·15 = 2700 mm², y_f = 280 + 15/2 = 287.5 mm

web: A_w = 8·280 = 2240 mm², y_w = 280/2 = 140 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2700·287.5 + 2240·140)/4940 = 220.6 mm

Parallel-axis theorem

flange: 180·15³/12 + 2700·(287.5 − 220.6)² = 1.213 × 10⁷ mm⁴

web: 8·280³/12 + 2240·(140 − 220.6)² = 2.919 × 10⁷ mm⁴

I_x = 1.213 × 10⁷ + 2.919 × 10⁷ = 4.132 × 10⁷ mm⁴ = 41.32 × 10⁶ mm⁴

Answer

I_x = 41.32 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 3.464 kN·m

Centroid

ȳ = (3900·187.5 + 1440·90)/5340 = 161.2 mm

top edge: c = 195 − 161.2 = 33.79 mm

Second moment of area

I = 260·15³/12 + 3900·26.29² + 8·180³/12 + 1440·71.21² = 1.396 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 90·1.396 × 10⁷/33.79 = 37.18 kN·m

bottom edge: M = 40·1.396 × 10⁷/161.2 = 3.464 kN·m

Governing value

M_allow = min = 3.464 kN·m (bottom edge)

Answer

M_allow = 3.464 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_max = 85.44 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 = √(((100 − −60)/2)² + −30²) = 85.44 MPa

Answer

τ_max = 85.44 MPa

10.Mohr’s circle & stress transformation

Answer: σ₁ = 48.19 MPa

Stresses at the surface

σ = F/A = −150000/(π·20²/4) = −150000/314.2 = −477.5 MPa

τ = 16T/(πd³) = 16·250000/(π·20³) = 159.2 MPa

Mohr’s circle

C = σ/2 = −238.7 MPa

R = √((−238.7)² + 159.2²) = 286.9 MPa

σ₁ = C + R = 48.19 MPa

σ₂ = C − R = −525.7 MPa

Answer

σ₁ = 48.19 MPa

11.Beam deflection

Answer: δ_B = 3.125 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 = 15000 N, a = 500 mm, L = 3000 mm

EI = 1.7 × 10¹² N·mm²

Deflection at the free end

Deflection at the load plus the slope at the load times the remaining length.

δ_P = Pa³/(3EI) = 0.3676 mm

θ_P = Pa²/(2EI) = 0.001103 rad

δ_B = δ_P + θ_P·(L − a) = Pa²(3L − a)/(6EI) = 3.125 mm

Answer

δ_B = 3.125 mm

12.Beam deflection

Answer: I_req = 920.5 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 6000/360 = 16.67 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·10·6000⁴/(384·11000·16.67) = 9.205 × 10⁸ mm⁴

I_req = 920.5 × 10⁶ mm⁴

Answer

I_req = 920.5 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 154.3 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 = 2·1500 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/3000² = 154342 N = 154.3 kN

Answer

P_cr = 154.3 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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