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 bar is made of a steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 mm, length 300 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 300 mm, E = 70 GPaδ = 0.2 mm

P = kN

2.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø100/Ø60 mm, E = 120 GPa, σ_allow = 100 MPa

P_max = kN

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 = 250 N·md = 20 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 3000 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 30 kWN = 3000 rpmτ_allow = 60 MPa

d = mm

5.Shear force & bending momentMedium

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. It carries a point load P₁ between the supports and a point load P₂ at C. Find the bending moment M_x under the load P₁. Use sagging moments positive and upward shear on the left face positive.

L = 6.5 mP₁ = 80 kN at 3 mP₂ = 70 kN at 6.5 msupports at x = 0 and x = 5 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.

30 kN @ 6 m8 kN/m, 0–5 m

1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 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 = 100 mmt_f = 30 mmt_w = 10 mmh_w = 220 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 30 MPa in tension and 120 MPa in compression. Find the largest hogging bending moment M_allow the beam can carry (give its size).

b_f = 180 mmt_f = 15 mmt_w = 20 mmh_w = 280 mmσ_t,allow = 30 MPaσ_c,allow = 120 MPa

M_allow = kN·m

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −20 MPa, σᵧ = 70 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = −20 MPaσ_y = 70 MPaτ_xy = 30 MPa

τ_max = MPa

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 8 mm holds gas at a gauge pressure of 1.2 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 500 mmt = 8 mmp = 1.2 MPaα = 20°

|τ_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 = 8 kN/mL = 4 mE = 70 GPa (aluminium)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 point load P at midspan. EI is constant. Find the reaction R_B at the prop.

L = 4 mP = 30 kN at 2 m

R_B = kN

13.Column bucklingMedium

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

L = 2 mE = 200 GPa (steel)b × h = 20 × 80 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 140 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 140 kNn = 2L = 4 mE = 70 GPa (aluminium)

d = mm

3Solutions

1.Axial stress & strain

Answer: P = 54.98 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 = 706.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(70000·2827) = 1.516 × 10⁻⁶ mm/N

Load

P = 0.2/3.638 × 10⁻⁶ = 54978 N = 54.98 kN

Answer

P = 54.98 kN

2.Axial stress & strain

Answer: P_max = 620.5 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·5027 = 7.446 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0008333) = 0.0008333 (copper)

Largest load

P_max = 0.0008333·7.446 × 10⁸ = 620465 N = 620.5 kN

Answer

P_max = 620.5 kN

3.Torsion of shafts

Answer: φ = 23.38 °

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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·1000/(39000·15708) = 0.4081 rad

φ = 0.4081·180/π = 23.38°

Answer

φ = 23.38°

4.Torsion of shafts

Answer: d = 20.09 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·60)) = 20.09 mm

Answer

d = 20.09 mm

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

5.Shear force & bending moment

Answer: M_x = 33 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. The overhang load also acts about A.

ΣM_A = 0: R_B·5 = 80·3 + 70·6.5

R_B = 695/5 = 139 kN

ΣF_y = 0: R_A = 150 − 139 = 11 kN

Cut under P₁ and keep the left part

M_x = R_A·3 = 11·3 = 33 kN·m

Answer

M_x = 33 kN·m

6.Shear force & bending moment

Answer: F (x = 5 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 = 14 kN, R₂ = 56 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 = 10, M_C = 12, M_D = 6, M_E = −8, M_F = −30 kN·m, M_G = 0

Answer

Mark F: M_F = −30 kN·m

7.Centroid & second moment of area

Answer: I_x = 28.93 × 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 = 100·30 = 3000 mm², y_f = 220 + 30/2 = 235 mm

web: A_w = 10·220 = 2200 mm², y_w = 220/2 = 110 mm

Centroid

ȳ = (A_fy_f + A_wy_w)/(A_f + A_w) = (3000·235 + 2200·110)/5200 = 182.1 mm

Parallel-axis theorem

flange: 100·30³/12 + 3000·(235 − 182.1)² = 8.615 × 10⁶ mm⁴

web: 10·220³/12 + 2200·(110 − 182.1)² = 2.031 × 10⁷ mm⁴

I_x = 8.615 × 10⁶ + 2.031 × 10⁷ = 2.893 × 10⁷ mm⁴ = 28.93 × 10⁶ mm⁴

Answer

I_x = 28.93 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: M_allow = 21.38 kN·m

Centroid

ȳ = (2700·287.5 + 5600·140)/8300 = 188 mm

top edge: c = 295 − 188 = 107 mm

Second moment of area

I = 180·15³/12 + 2700·99.52² + 20·280³/12 + 5600·47.98² = 7.627 × 10⁷ mm⁴

Moment allowed by each edge

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

top edge: M = 30·7.627 × 10⁷/107 = 21.38 kN·m

bottom edge: M = 120·7.627 × 10⁷/188 = 48.69 kN·m

Governing value

M_allow = min = 21.38 kN·m (top edge)

Answer

M_allow = 21.38 kN·m

9.Mohr’s circle & stress transformation

Answer: τ_max = 54.08 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.

Radius of Mohr’s circle

τ_max = R = √(((−20 − 70)/2)² + 30²) = 54.08 MPa

Answer

τ_max = 54.08 MPa

10.Mohr’s circle & stress transformation

Answer: |τ_n| = 6.026 MPa

Wall stresses

r = 500/2 = 250 mm

σ_h = pr/t = 1.2·250/8 = 37.5 MPa

σ_a = pr/(2t) = 18.75 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.

|τ_n| = (37.5 − 18.75)·sin 20°·cos 20° = 6.026 MPa

Answer

|τ_n| = 6.026 MPa

11.Beam deflection

Answer: δ_max = 16.56 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 = 8 N/mm, L = 4000 mm

EI = 1.61 × 10¹² N·mm²

Deflection

δ_max = 5wL⁴/(384EI) = 5·8·4000⁴/(384·1.61 × 10¹²) = 16.56 mm

Answer

δ_max = 16.56 mm

12.Beam deflection

Answer: R_B = 9.375 kN

Remove the prop

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

δ_load = P(L/2)²(3L − L/2)/(6EI) = 5PL³/(48EI)

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) = 5PL³/(48EI) → R_B = 5P/16 = 5·30/16 = 9.375 kN

Answer

R_B = 9.375 kN

13.Column buckling

Answer: P_cr = 6.58 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 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·53333/4000² = 6580 N = 6.58 kN

Answer

P_cr = 6.58 kN

14.Column buckling

Answer: d = 89.7 mm

Required critical load

P_cr = n·P = 2·140000 = 280000 N

KL = 0.7·4000 = 2800 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·280000·2800²/(π³·70000))^(1/4) = 89.7 mm

Answer

d = 89.7 mm

Check: the slenderness KL/r is 124.9, so the column is slender and Euler applies.

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

Mechanics of Materials: cram pack

1Formulas and methods

Axial stress & strain

Torsion of shafts

Method

Shear force & bending moment

Method

Centroid & second moment of area

Mohr’s circle & stress transformation

Method

Beam deflection

Column buckling

2Practice problems

3Solutions

Areas

Flexibilities

Load

Answer

Areas

Combined axial stiffness

Governing strain

Largest load

Answer

Polar second moment of area

Angle of twist

Answer

Torque

Diameter from the allowable stress

Answer

Support reactions

Cut under P1 and keep the left part

Answer

Support reactions

Bending moment at every mark

Answer

Split into two rectangles

Centroid

Parallel-axis theorem

Answer

Centroid

Second moment of area

Moment allowed by each edge

Governing value

Answer

Radius of Mohr’s circle

Answer

Wall stresses

Transform to the seam

Answer

Consistent units

Deflection

Answer

Remove the prop

Compatibility at the prop

Answer

Smallest second moment of area

Effective length

Critical load

Answer

Required critical load

Solve for the diameter

Answer

Cut under P₁ and keep the left part