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 aluminium bar under the axial load.

P = 110 kNd = 25 mmL = 2.5 mE = 70 GPa

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 400 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 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: bronze, Ø30 mm, 400 mm, E = 110 GPaP = 20 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid aluminium 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 = 100 N·md = 25 mmL = 1.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 30 kW at 500 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter d_o.

P = 30 kWN = 500 rpmd_i/d_o = 0.5τ_allow = 60 MPa

d_o = 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.

20 kN @ 6 m10 kN/m, 0–12 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)

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 distance x_c from A to the point of contraflexure between the supports (where the bending moment changes sign).

L = 7.5 mw = 5 kN/m from 0 to 7.5 msupports at x = 0 and x = 6 m

x_c = m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 130 mm and inside diameter d = 106 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 130 mmd = 106 mm

r = mm

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 hogging bending moment M_allow the beam can carry (give its size).

b_f = 280 mmt_f = 25 mmt_w = 20 mmh_w = 220 mmσ_t,allow = 40 MPaσ_c,allow = 90 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 60 MPa, σᵧ = 100 MPa and τₓᵧ = 60 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 60 MPaσ_y = 100 MPaτ_xy = 60 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −60 MPa, σᵧ = −20 MPa and τₓᵧ = −50 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −60 MPaσ_y = −20 MPaτ_xy = −50 MPaθ = 20°

τ_θ = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 2.5 m from support A. Find the deflection δ_P under the load.

P = 7 kNa = 2.5 mL = 4 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ_P = 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 reaction R_B at the prop.

L = 4 mw = 12 kN/m

R_B = kN

13.Column bucklingMedium

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

L = 1.5 mE = 70 GPa (aluminium)b × h = 40 × 60 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 145 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 145 kNn = 2L = 6 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 8.003 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·25²/4 = 490.9 mm²

σ = P/A = 110·1000/490.9 = 224.1 MPa

Strain (Hooke’s law)

E for aluminium = 70 GPa = 70000 MPa.

ε = σ/E = 224.1/70000 = 0.003201 = 3201 µε

Elongation

δ = ε·L = 0.003201·2500 mm = 8.003 mm

Answer

δ = 8.003 mm

2.Axial stress & strain

Answer: σ₂ = 8.968 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 = 200000·1257/600 = 418879 N/mm

k₂ = 110000·706.9/400 = 194386 N/mm

Share of segment 2

F₂ = 20000·194386/613265 = 6339 N

Stress

σ₂ = 6339/706.9 = 8.968 MPa (compression)

Answer

σ₂ = 8.968 MPa

3.Torsion of shafts

Answer: φ = 8.619 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1500/(26000·38350) = 0.1504 rad

φ = 0.1504·180/π = 8.619°

Answer

φ = 8.619°

4.Torsion of shafts

Answer: d_o = 37.3 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 30000/52.36 = 573 N·m

Outer diameter from the allowable stress

T = 573 N·m = 572958 N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·572958/(π·60·0.9375)) = 37.3 mm

d_i = 0.5·37.3 = 18.65 mm

Answer

d_o = 37.3 mm

A solid shaft would need d = 36.5 mm; the hollow one is a little larger but much lighter.

5.Shear force & bending moment

Answer: D (x = 6 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 = 70 kN, R₂ = 70 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, M_C = 200, M_D = 240 kN·m, M_E = 200, M_F = 120, M_G = 0

Answer

Mark D: M_D = 240 kN·m

6.Shear force & bending moment

Answer: x_c = 5.625 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 = 5·7.5 = 37.5 kN, acting at x̄ = 3.75 m

ΣM_A = 0: R_B·6 = 5·7.5·3.75

R_B = 140.6/6 = 23.44 kN

ΣF_y = 0: R_A = 37.5 − 23.44 = 14.06 kN

Zero moment

Between the supports M(x) = R_A·x − w·x²/2. Apart from x = 0 it is zero at x = 2R_A/w.

R_A·x_c − w·x_c²/2 = 0

x_c = 2·14.06/5 = 5.625 m

Answer

x_c = 5.625 m

7.Centroid & second moment of area

Answer: r = 41.93 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 4448 mm²

I = π(D⁴ − d⁴)/64 = 7.823 × 10⁶ mm⁴

Radius of gyration

r = √(I/A) = √(7.823 × 10⁶/4448) = 41.93 mm

Answer

r = 41.93 mm

8.Centroid & second moment of area

Answer: M_allow = 28.5 kN·m

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.

Centroid

ȳ = (7000·232.5 + 4400·110)/11400 = 185.2 mm

top edge: c = 245 − 185.2 = 59.78 mm

Second moment of area

I = 280·25³/12 + 7000·47.28² + 20·220³/12 + 4400·75.22² = 5.865 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 40·5.865 × 10⁷/59.78 = 39.25 kN·m

bottom edge: M = 90·5.865 × 10⁷/185.2 = 28.5 kN·m

Governing value

M_allow = min = 28.5 kN·m (bottom edge)

Answer

M_allow = 28.5 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_abs = 71.62 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 80 MPa, R = √((−20)² + 60²) = 63.25 MPa

σ₁ = 143.2 MPa, σ₂ = 16.75 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(126.5, 143.2, 16.75)/2 = 71.62 MPa

Answer

τ_abs = 71.62 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: τ_θ = −25.45 MPa

Transformation equation

τ_θ = −−20·0.6428 + −50·0.766 = −25.45 MPa

Answer

τ_θ = −25.45 MPa

11.Beam deflection

Answer: δ_P = 1.395 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 = 7000 N, a = 2500 mm, b = 1500 mm, L = 4000 mm

EI = 5.88 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 7000·2500²·1500²/(3·5.88 × 10¹²·4000) = 1.395 mm

Answer

δ_P = 1.395 mm

12.Beam deflection

Answer: R_B = 18 kN

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 = 3·12·4/8 = 18 kN

Answer

R_B = 18 kN

13.Column buckling

Answer: P_cr = 200.5 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 = 60·40³/12 = 320000 mm⁴

Effective length

KL = 0.7·1500 = 1050 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·320000/1050² = 200525 N = 200.5 kN

Answer

P_cr = 200.5 kN

14.Column buckling

Answer: d = 85.24 mm

Required critical load

P_cr = n·P = 2·145000 = 290000 N

KL = 0.7·6000 = 4200 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·290000·4200²/(π³·200000))^(1/4) = 85.24 mm

Answer

d = 85.24 mm

Check: the slenderness KL/r is 197.1, so the column is slender and Euler applies.

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