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 30 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 600 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø30 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 600 mm

δ = mm

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 = 115 kNsegment 1: Ø40 mm, 1 m, copper (E = 117 GPa)segment 2: Ø20 mm, 1.5 m, brass (E = 100 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 = 400 N·md = 50 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 1500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 20 kWN = 1500 rpmτ_allow = 50 MPa

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 = 4 mw = 5 kN/m from 0 to 4 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.

60 kN @ 0 m55 kN @ 12 m45 kN @ 8 m5 kN/m, 2–10 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 channel section lies with its open side up: a base plate 300 mm wide with two upright legs, 100 mm high overall. Base and legs are all 8 mm thick. Find its second moment of area I_x about the horizontal axis x through its centroid.

b = 300 mmh = 100 mmt = 8 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 = 200 mmt_f = 25 mmt_w = 16 mmh_w = 240 mmM = 30 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 130 MPa, σᵧ = 40 MPa and τₓᵧ = 30 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 = 130 MPaσ_y = 40 MPaτ_xy = 30 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 40 MPa, σᵧ = 90 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 = 40 MPaσ_y = 90 MPaτ_xy = 40 MPaθ = -45°

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 = 15 kN/mL = 4 mE = 200 GPa (steel)I = 23 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the size of the bending moment M_A at the fixed end.

L = 3 mw = 8 kN/m

M_A = kN·m

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 = 3 mE = 200 GPa (steel)b × h = 30 × 100 mm

P_cr = kN

14.Column bucklingHard

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

L = 4 mE = 200 GPa (steel)d = 50 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: δ = 0.03801 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·30²/4 = 706.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·5890 = 7.893 × 10⁸ N

Shortening

δ = 50000·600/7.893 × 10⁸ = 0.03801 mm

Answer

δ = 0.03801 mm

2.Axial stress & strain

Answer: σ_max = 366.1 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·40²/4 = 1257 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 91.51 MPa

σ₂ = 366.1 MPa

Largest stress

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

σ_max = 366.1 MPa

Answer

σ_max = 366.1 MPa

3.Torsion of shafts

Answer: φ = 0.9338 °

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) = 400·1000·2000/(80000·613592) = 0.0163 rad

φ = 0.0163·180/π = 0.9338°

Answer

φ = 0.9338°

4.Torsion of shafts

Answer: d = 23.49 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 20000/157.1 = 127.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·127324/(π·50)) = 23.49 mm

Answer

d = 23.49 mm

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

5.Shear force & bending moment

Answer: M_x = 7.5 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 = 5·4 = 20 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 5·4·2

R_B = 40/4 = 10 kN

ΣF_y = 0: R_A = 20 − 10 = 10 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 = 10·1 − 5·1·0.5 = 7.5 kN·m

Answer

M_x = 7.5 kN·m

6.Shear force & bending moment

Answer: B (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 = 92.5 kN, R₂ = 107.5 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 = −120 kN·m, M_C = −65, M_D = −30, M_E = −15, M_F = −110, M_G = 0

Answer

Mark B: M_B = −120 kN·m

7.Centroid & second moment of area

Answer: I_x = 3.332 × 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 rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 300·8 = 2400 mm², y₁ = 4 mm

two legs: A₂ = 2·8·92 = 1472 mm², y₂ = 8 + 92/2 = 54 mm

Centroid

ȳ = (2400·4 + 1472·54)/3872 = 23.01 mm

Parallel-axis theorem

base: 300·8³/12 + 2400·(4 − 23.01)² = 879954 mm⁴

two legs: 2·8·92³/12 + 1472·(54 − 23.01)² = 2.452 × 10⁶ mm⁴

I_x = 879954 + 2.452 × 10⁶

I_x = 3.332 × 10⁶ mm⁴ = 3.332 × 10⁶ mm⁴

Answer

I_x = 3.332 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 102.9 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 200·25 = 5000 mm², y_f = 240 + 25/2 = 252.5 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) = (5000·252.5 + 3840·120)/8840 = 194.9 mm

Parallel-axis theorem

flange: 200·25³/12 + 5000·(252.5 − 194.9)² = 1.682 × 10⁷ mm⁴

web: 16·240³/12 + 3840·(120 − 194.9)² = 4 × 10⁷ mm⁴

I_x = 1.682 × 10⁷ + 4 × 10⁷ = 5.682 × 10⁷ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 194.9 mm

σ = M·y/I = 3 × 10⁷·194.9/5.682 × 10⁷ = 102.9 MPa

Answer

σ_bottom = 102.9 MPa

9.Mohr’s circle & stress transformation

Answer: θ_p = 16.85 °

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(2·30, 130 − 40) = 33.69°

Angle on the element

θ_p = 16.85°

Answer

θ_p = 16.85°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_θ = 25 MPa

Transformation equation

σ_θ = 65 + −25·0 + 40·−1 = 25 MPa

Answer

σ_θ = 25 MPa

11.Beam deflection

Answer: δ_max = 10.87 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 = 15 N/mm, L = 4000 mm

EI = 4.6 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·15·4000⁴/(384·4.6 × 10¹²) = 10.87 mm

Answer

δ_max = 10.87 mm

12.Beam deflection

Answer: M_A = 9 kN·m

Remove the prop

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

δ_load = wL⁴/(8EI)

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) = wL⁴/(8EI) → R_B = 3wL/8 = 3·8·3/8 = 9 kN

Moment at the fixed end

Moments about A of the loads and the prop force.

M_A = wL²/2 − R_B·L = 36 − 27 = 9 kN·m

Answer

M_A = 9 kN·m

13.Column buckling

Answer: P_cr = 100.7 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 = 100·30³/12 = 225000 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·225000/2100² = 100710 N = 100.7 kN

Answer

P_cr = 100.7 kN

14.Column buckling

Answer: σ_cr = 77.11 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 0.5·4000 = 2000 mm

KL/r = 2000/12.5 = 160

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/160² = 77.11 MPa

Answer

σ_cr = 77.11 MPa

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