Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 25 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 GPa). Both are 600 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø25 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 100 GPaL = 600 mm

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 60 mm, length 400 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 400 mm, E = 70 GPaP = 30 kN

σ₂ = MPa

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 250 mm rotate 1.08° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 15 mmL = 250 mmφ = 1.08°

G = GPa

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 1450 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 5 kWN = 1450 rpmτ_allow = 75 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.

45 kN @ 4 m4 kN/m, 2–12 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 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 = 4 mP = 10 kN at 2 mw = 2 kN/m from 0 to 4 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 120 mm × 150 mm plate has a circular hole of diameter 40 mm on its vertical centre line, with its centre 45 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 120 mmh = 150 mmd = 40 mmhole centre: y = 45 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 20 kN·m. Find the bending stress σ_bottom at the bottom edge.

b_f = 120 mmt_f = 25 mmt_w = 20 mmh_w = 100 mmM = 20 kN·m

σ_bottom = MPa

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 150 MPa, σᵧ = 40 MPa and τₓᵧ = −40 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 150 MPaσ_y = 40 MPaτ_xy = −40 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 6 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1000 mmt = 6 mmp = 3 MPaα = 55°

σ_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 = 5 mE = 70 GPa (aluminium)I = 12 × 10⁶ mm⁴

δ_max = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 75 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth h_req of the section.

w = 2 kN/mL = 5 mE = 11 GPa (timber)b = 75 mm

h_req = mm

13.Column bucklingMedium

A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 200 GPa (steel)d = 30 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 80 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 80 kNn = 2.5L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.1187 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·25²/4 = 490.9 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·3063 = 4.045 × 10⁸ N

Shortening

δ = 80000·600/4.045 × 10⁸ = 0.1187 mm

Answer

δ = 0.1187 mm

2.Axial stress & strain

Answer: σ₂ = 7.974 MPa

Stiffnesses

k₁ = 200000·490.9/600 = 163625 N/mm

k₂ = 70000·2827/400 = 494801 N/mm

Share of segment 2

F₂ = 30000·494801/658425 = 22545 N

Stress

σ₂ = 22545/2827 = 7.974 MPa (compression)

Answer

σ₂ = 7.974 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 = 1.08·π/180 = 0.01885 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·250/(4970·0.01885) = 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: d = 13.08 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 5000/151.8 = 32.93 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·32929/(π·75)) = 13.08 mm

Answer

d = 13.08 mm

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

5.Shear force & bending moment

Answer: C (x = 4 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 = 46.67 kN, R₂ = 38.33 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 = 93.33, M_C = 178.7 kN·m, M_D = 158, M_E = 121.3, M_F = 68.67, M_G = 0

Answer

Mark C: M_C = 178.7 kN·m

6.Shear force & bending moment

Answer: M_max = 14 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 = 2·4 = 8 kN, acting at x̄ = 2 m

ΣM_A = 0: R_B·4 = 10·2 + 2·4·2

R_B = 36/4 = 9 kN

ΣF_y = 0: R_A = 18 − 9 = 9 kN

Where is V = 0?

V(2⁻) = R_A − w·2 = 5 kN

V(2⁺) = 5 − 10 = −5 kN

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

Maximum moment

M_max = R_A·2 − w·2²/2 = 9·2 − 2·2²/2 = 14 kN·m

Answer

M_max = 14 kN·m

It occurs at x = 2 m from A.

7.Centroid & second moment of area

Answer: I_x = 32.41 × 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 = (18000·75 − 1257·45)/(18000 − 1257) = 77.25 mm

Full plate about ȳ

120·150³/12 + 18000·−2.252² = 33.84 × 10⁶ mm⁴

Hole about ȳ

π·40⁴/64 + 1257·−32.25² = 1.433 × 10⁶ mm⁴

Subtract

I_x = 32.41 × 10⁶ mm⁴

Answer

I_x = 32.41 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_bottom = 268.8 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 = 120·25 = 3000 mm², y_f = 100 + 25/2 = 112.5 mm

web: A_w = 20·100 = 2000 mm², y_w = 100/2 = 50 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3000·112.5 + 2000·50)/5000 = 87.5 mm

Parallel-axis theorem

flange: 120·25³/12 + 3000·(112.5 − 87.5)² = 2.031 × 10⁶ mm⁴

web: 20·100³/12 + 2000·(50 − 87.5)² = 4.479 × 10⁶ mm⁴

I_x = 2.031 × 10⁶ + 4.479 × 10⁶ = 6.51 × 10⁶ mm⁴

Bending stress

The bottom edge is ȳ from the axis.

y = 87.5 mm

σ = M·y/I = 2 × 10⁷·87.5/6.51 × 10⁶ = 268.8 MPa

Answer

σ_bottom = 268.8 MPa

9.Mohr’s circle & stress transformation

Answer: τ_abs = 81.5 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 = 95 MPa, R = √(55² + (−40)²) = 68.01 MPa

σ₁ = 163 MPa, σ₂ = 26.99 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(136, 163, 26.99)/2 = 81.5 MPa

Answer

τ_abs = 81.5 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: σ_n = 166.1 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. Thin-walled cylinder: hoop stress σ_h = pr/t and axial stress σ_a = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 3·500/6 = 250 MPa

σ_a = pr/(2t) = 125 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.

σ_n = 125·sin²55° + 250·cos²55° = 166.1 MPa

Answer

σ_n = 166.1 MPa

11.Beam deflection

Answer: δ_max = 38.75 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 = 5000 mm

EI = 8.4 × 10¹¹ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·4·5000⁴/(384·8.4 × 10¹¹) = 38.75 mm

Answer

δ_max = 38.75 mm

12.Beam deflection

Answer: h_req = 257.4 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.

w = 2 N/mm, L = 5000 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 5000/360 = 13.89 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·2·5000⁴/(384·11000·13.89) = 1.065 × 10⁸ mm⁴

Required depth

bh³/12 = I_req → h = (12·1.065 × 10⁸/75)^1/3 = 257.4 mm

Answer

h_req = 257.4 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 19.62 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·39761/2000² = 19621 N = 19.62 kN

Answer

P_cr = 19.62 kN

14.Column buckling

Answer: d = 63.43 mm

Required critical load

P_cr = n·P = 2.5·80000 = 200000 N

KL = 0.7·4000 = 2800 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·200000·2800²/(π³·200000))^(1/4) = 63.43 mm

Answer

d = 63.43 mm

Check: the slenderness KL/r is 176.6, so the column is slender and Euler applies.

ExamProblems · examproblems.com/courses/mechanics-of-materials