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 30 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 60 mm, E = 70 GPa). Both are 400 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.

P = 200 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 70 GPaL = 400 mm

δ = mm

2.Axial stress & strainHard

A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 500 mm, E = 70 GPaP = 80 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A solid brass 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 = 25 mmL = 0.8 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 3000 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 30 kWN = 3000 rpmτ_allow = 50 MPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 2.5 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 5 mw = 10 kN/m from 0 to 5 mx = 2.5 m

M_x = kN·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 = 70 kN at 4 mw = 5 kN/m from 0 to 10 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A 150 mm × 150 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 97.5 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 150 mmd = 30 mmhole centre: y = 97.5 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 40 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 280 mmt_f = 20 mmt_w = 10 mmh_w = 260 mmM = 40 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 30 MPa and σᵧ = −40 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 65 MPa?

σ_x = 30 MPaσ_y = −40 MPaσ₁ = 65 MPa

τ_xy = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −30 MPa, σᵧ = 60 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.

σ_x = −30 MPaσ_y = 60 MPaτ_xy = −60 MPaθ = 40°

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 = 10 kN/mL = 5 mE = 200 GPa (steel)I = 160 × 10⁶ mm⁴

δ_max = 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 = 20 kN/mL = 1.5 mE = 200 GPa (steel)

I_req = × 10⁶ mm⁴

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 = 2.5 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 25 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 25 kNn = 3L = 4 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: δ = 0.2709 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(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·2199 = 2.953 × 10⁸ N

Shortening

δ = 200000·400/2.953 × 10⁸ = 0.2709 mm

Answer

δ = 0.2709 mm

2.Axial stress & strain

Answer: σ₂ = 21.16 MPa

Stiffnesses

k₁ = 200000·706.9/400 = 353429 N/mm

k₂ = 70000·1257/500 = 175929 N/mm

Share of segment 2

F₂ = 80000·175929/529358 = 26588 N

Stress

σ₂ = 26588/1257 = 21.16 MPa (compression)

Answer

σ₂ = 21.16 MPa

3.Torsion of shafts

Answer: φ = 12.26 °

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) = 400·1000·800/(39000·38350) = 0.214 rad

φ = 0.214·180/π = 12.26°

Answer

φ = 12.26°

4.Torsion of shafts

Answer: d = 21.35 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 30000/314.2 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·50)) = 21.35 mm

Answer

d = 21.35 mm

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

5.Shear force & bending moment

Answer: M_x = 31.25 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 10·5 = 50 kN, acting at x̄ = 2.5 m

ΣM_A = 0: R_B·5 = 10·5·2.5

R_B = 125/5 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 2.5 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 = 25·2.5 − 10·2.5·1.25 = 31.25 kN·m

Answer

M_x = 31.25 kN·m

6.Shear force & bending moment

Answer: M_max = 228 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 = 5·10 = 50 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 70·4 + 5·10·5

R_B = 530/10 = 53 kN

ΣF_y = 0: R_A = 120 − 53 = 67 kN

Where is V = 0?

V(4⁻) = R_A − w·4 = 47 kN

V(4⁺) = 47 − 70 = −23 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 = 67·4 − 5·4²/2 = 228 kN·m

Answer

M_max = 228 kN·m

It occurs at x = 4 m from A.

7.Centroid & second moment of area

Answer: I_x = 41.78 × 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 = (22500·75 − 706.9·97.5)/(22500 − 706.9) = 74.27 mm

Full plate about ȳ

150·150³/12 + 22500·0.7298² = 42.2 × 10⁶ mm⁴

Hole about ȳ

π·30⁴/64 + 706.9·23.23² = 0.4212 × 10⁶ mm⁴

Subtract

I_x = 41.78 × 10⁶ mm⁴

Answer

I_x = 41.78 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 43.83 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 = 280·20 = 5600 mm², y_f = 260 + 20/2 = 270 mm

web: A_w = 10·260 = 2600 mm², y_w = 260/2 = 130 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (5600·270 + 2600·130)/8200 = 225.6 mm

Parallel-axis theorem

flange: 280·20³/12 + 5600·(270 − 225.6)² = 1.122 × 10⁷ mm⁴

web: 10·260³/12 + 2600·(130 − 225.6)² = 3.841 × 10⁷ mm⁴

I_x = 1.122 × 10⁷ + 3.841 × 10⁷ = 4.964 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 280 − 225.6 = 54.39 mm

σ = M·y/I = 4 × 10⁷·54.39/4.964 × 10⁷ = 43.83 MPa

Answer

σ_top = 43.83 MPa

9.Mohr’s circle & stress transformation

Answer: τ_xy = 60.62 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.

Required radius

C = (30 + (−40))/2 = −5 MPa

R = σ₁ − C = 65 − (−5) = 70 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(70² − 35²) = 60.62 MPa

Answer

τ_xy = 60.62 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = −51.9 MPa

Transformation equation

σ_θ = 15 + −45·0.1736 + −60·0.9848 = −51.9 MPa

Answer

σ_θ = −51.9 MPa

11.Beam deflection

Answer: δ_max = 2.543 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 = 10 N/mm, L = 5000 mm

EI = 3.2 × 10¹³ N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·10·5000⁴/(384·3.2 × 10¹³) = 2.543 mm

Answer

δ_max = 2.543 mm

12.Beam deflection

Answer: I_req = 1.582 × 10⁶ mm⁴

Deflection limit

δ_allow = L/360 = 1500/360 = 4.167 mm

Required second moment of area

I = 5wL⁴/(384·E·δ_allow) = 5·20·1500⁴/(384·200000·4.167) = 1.582 × 10⁶ mm⁴

I_req = 1.582 × 10⁶ mm⁴

Answer

I_req = 1.582 × 10⁶ mm⁴

13.Column buckling

Answer: P_cr = 28.35 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·2500 = 1750 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/1750² = 28349 N = 28.35 kN

Answer

P_cr = 28.35 kN

14.Column buckling

Answer: d = 59.32 mm

Required critical load

P_cr = n·P = 3·25000 = 75000 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·75000·4000²/(π³·200000))^(1/4) = 59.32 mm

Answer

d = 59.32 mm

Check: the slenderness KL/r is 269.7, so the column is slender and Euler applies.

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