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 bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 200 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, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 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 total elongation δ.

P = 35 kNsegment 1: Ø30 mm, 1 m, steel (E = 200 GPa)segment 2: Ø16 mm, 2 m, aluminium (E = 70 GPa)

δ = mm

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 = 200 N·md = 50 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 30 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 30 kWN = 300 rpmd_o = 100 mmd_i = 60 mm

τ_max = MPa

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.

30 kN @ 6 m5 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

Find the bending moment M_x at the section x = 2 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mP = 30 kN at 2 mw = 8 kN/m from 0 to 8 mx = 2 m

M_x = 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 = 140 mmt_f = 15 mmt_w = 14 mmh_w = 200 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 25 kN·m. Find the size of the bending stress σ_top at the top edge.

b_f = 160 mmt_f = 30 mmt_w = 18 mmh_w = 120 mmM = 25 kN·m

σ_top = MPa

9.Mohr’s circle & stress transformationMedium

A point has σₓ = 10 MPa, σᵧ = 0 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 35° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 10 MPaσ_y = 0 MPaτ_xy = 30 MPaθ = 35°

σ_y′ = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −40 MPa, σᵧ = 20 MPa and τₓᵧ = −50 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 = 20 MPaτ_xy = −50 MPaθ = -45°

σ_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 = 4 kNa = 2.5 mL = 4 mE = 70 GPa (aluminium)I = 36 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A simply supported beam carries a uniformly distributed load w and a point load P at midspan. Using superposition, find the midspan deflection δ_max.

P = 3 kNw = 6 kN/mL = 5 mE = 200 GPa (steel)I = 8.5 × 10⁶ mm⁴

δ_max = mm

13.Column bucklingMedium

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

L = 2.5 mE = 70 GPa (aluminium)d = 50 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 = 3 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 37.7 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 = 314.2 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 200/(100000·1257) = 1.592 × 10⁻⁶ mm/N

Load

P = 0.3/7.958 × 10⁻⁶ = 37699 N = 37.7 kN

Answer

P = 37.7 kN

2.Axial stress & strain

Answer: δ = 5.221 mm

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·16²/4 = 201.1 mm²

σ₁ = 49.51 MPa

σ₂ = 174.1 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 49.51·1000/200000 = 0.2476 mm

δ₂ = σ₂L₂/E₂ = 174.1·2000/70000 = 4.974 mm

δ = δ₁ + δ₂ = 5.221 mm

Answer

δ = 5.221 mm

3.Torsion of shafts

Answer: φ = 0.4669 °

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) = 200·1000·2000/(80000·613592) = 0.008149 rad

φ = 0.008149·180/π = 0.4669°

Answer

φ = 0.4669°

4.Torsion of shafts

Answer: τ_max = 5.588 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 30000/31.42 = 954.9 N·m

Polar second moment of area

J = π(100⁴ − 60⁴)/32 = 8.545 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 954930·50/8.545 × 10⁶ = 5.588 MPa

Answer

τ_max = 5.588 MPa

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 = 45 kN, R₂ = 45 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 = 80, M_C = 140, M_D = 180 kN·m, M_E = 140, M_F = 80, M_G = 0

Answer

Mark D: M_D = 180 kN·m

6.Shear force & bending moment

Answer: M_x = 93 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 = 8·8 = 64 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 30·2 + 8·8·4

R_B = 316/8 = 39.5 kN

ΣF_y = 0: R_A = 94 − 39.5 = 54.5 kN

Cut at x = 2 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 = 54.5·2 − 8·2·1 = 93 kN·m

Answer

M_x = 93 kN·m

7.Centroid & second moment of area

Answer: I_x = 23.24 × 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 = 140·15 = 2100 mm², y_f = 200 + 15/2 = 207.5 mm

web: A_w = 14·200 = 2800 mm², y_w = 200/2 = 100 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (2100·207.5 + 2800·100)/4900 = 146.1 mm

Parallel-axis theorem

flange: 140·15³/12 + 2100·(207.5 − 146.1)² = 7.964 × 10⁶ mm⁴

web: 14·200³/12 + 2800·(100 − 146.1)² = 1.528 × 10⁷ mm⁴

I_x = 7.964 × 10⁶ + 1.528 × 10⁷ = 2.324 × 10⁷ mm⁴ = 23.24 × 10⁶ mm⁴

Answer

I_x = 23.24 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: σ_top = 84.45 MPa

Split into two rectangles

Measure every y from the bottom edge.

flange: A_f = 160·30 = 4800 mm², y_f = 120 + 30/2 = 135 mm

web: A_w = 18·120 = 2160 mm², y_w = 120/2 = 60 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (4800·135 + 2160·60)/6960 = 111.7 mm

Parallel-axis theorem

flange: 160·30³/12 + 4800·(135 − 111.7)² = 2.96 × 10⁶ mm⁴

web: 18·120³/12 + 2160·(60 − 111.7)² = 8.371 × 10⁶ mm⁴

I_x = 2.96 × 10⁶ + 8.371 × 10⁶ = 1.133 × 10⁷ mm⁴

Bending stress

The top edge is h − ȳ from the axis.

y = 150 − 111.7 = 38.28 mm

σ = M·y/I = 2.5 × 10⁷·38.28/1.133 × 10⁷ = 84.45 MPa

Answer

σ_top = 84.45 MPa

9.Mohr’s circle & stress transformation

Answer: σ_y′ = −24.9 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.

Transformation equation for the y′ face

σ_y′ = 5 − 5·0.342 − 30·0.9397 = −24.9 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 34.9 MPa, σ_x′ + σ_y′ = 10 = σ_x + σ_y

Answer

σ_y′ = −24.9 MPa

10.Mohr’s circle & stress transformation

Answer: σ_θ = 40 MPa

Transformation equation

σ_θ = −10 + −30·0 + −50·−1 = 40 MPa

Answer

σ_θ = 40 MPa

11.Beam deflection

Answer: δ_P = 1.86 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 = 4000 N, a = 2500 mm, b = 1500 mm, L = 4000 mm

EI = 2.52 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 4000·2500²·1500²/(3·2.52 × 10¹²·4000) = 1.86 mm

Answer

δ_P = 1.86 mm

12.Beam deflection

Answer: δ_max = 33.32 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.

P = 3000 N, w = 6 N/mm, L = 5000 mm, EI = 1.7 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(48EI) = 3000·5000³/(48·1.7 × 10¹²) = 4.596 mm

δ_w = 5wL⁴/(384EI) = 5·6·5000⁴/(384·1.7 × 10¹²) = 28.72 mm

Add them

δ_max = 4.596 + 28.72 = 33.32 mm

Answer

δ_max = 33.32 mm

13.Column buckling

Answer: P_cr = 33.91 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 1·2500 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·306796/2500² = 33913 N = 33.91 kN

Answer

P_cr = 33.91 kN

14.Column buckling

Answer: σ_cr = 30.71 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/10 = 150

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/150² = 30.71 MPa

Answer

σ_cr = 30.71 MPa

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