Analysis of Bars of Varying Sections- Numerical Set 2

Analysis of Bars of Varying Sections

Many practical bars are made of different lengths and with different diameters, so their cross-sectional areas are not the same along the length. Figure 1.6(a) shows a bar composed of three segments, each with its own area and length. The entire bar is subjected to an axial load P.

Even though the same load acts through all sections, the stress, strain and extension in each part will be different because the areas (and possibly material properties) are different. The total change in length of the bar is obtained by adding the change in length of each segment.

Let:

• P = axial load on the bar
• L₁, L₂, L₃ = lengths of sections 1, 2 and 3
• A₁, A₂, A₃ = cross-sectional areas of sections 1, 2 and 3
• E = Young’s Modulus for the material of the bar (same for all sections here)

Stress in each section:

σ₁ = P / A₁     (section 1)
σ₂ = P / A₂     (section 2)
σ₃ = P / A₃     (section 3)

Strain in each section:

e₁ = σ₁ / E = P / (A₁ E)
e₂ = σ₂ / E = P / (A₂ E)
e₃ = σ₃ / E = P / (A₃ E)

Since strain is also equal to change in length divided by original length:

e₁ = ΔL₁ / L₁,   e₂ = ΔL₂ / L₂,   e₃ = ΔL₃ / L₃

Hence change in length of each section is:

ΔL₁ = (P L₁) / (A₁ E)
ΔL₂ = (P L₂) / (A₂ E)
ΔL₃ = (P L₃) / (A₃ E)

Total extension of the whole bar:

ΔL = ΔL₁ + ΔL₂ + ΔL₃ = P/E   [ L₁/A₁ + L₂/A₂ + L₃/A₃ ]

If the segments are made of different materials (different E values), the general form is:

ΔL = P  [ L₁/(A₁E₁) + L₂/(A₂E₂) + L₃/(A₃E₃) + … ]

Problem 1.8 – Stresses and Extension in a Bar of Three Sections

Question

An axial pull of 35,000 N acts on a bar made of three segments as shown in Fig. 1.6(b). The bar is of steel with Young’s Modulus E = 2.1 × 10⁵ N/mm². Determine:

1. (i) the stresses in each of the three sections, and
2. (ii) the total extension of the bar.
   

Data from Fig. 1.6(b)

Axial load, P = 35,000 N
Length of section 1, L₁ = 20 cm = 200 mm
Diameter of section 1, D₁ = 2 cm = 20 mm
Length of section 2, L₂ = 25 cm = 250 mm
Diameter of section 2, D₂ = 3 cm = 30 mm
Length of section 3, L₃ = 22 cm = 220 mm
Diameter of section 3, D₃ = 5 cm = 50 mm
Young’s Modulus, E = 2.1 × 10⁵ N/mm²

1. Cross-Sectional Areas

A₁ = (π/4) D₁² = (π/4)(20)² = 100π mm²
A₂ = (π/4) D₂² = (π/4)(30)² = 225π mm²
A₃ = (π/4) D₃² = (π/4)(50)² = 625π mm²

(i) Stresses in Each Section

Section 1:

σ₁ = P / A₁ = 35,000 / (100π) = 111.408 N/mm²

Section 2:

σ₂ = P / A₂ = 35,000 / (225π) = 49.5146 N/mm²

Section 3:

σ₃ = P / A₃ = 35,000 / (625π) = 17.8258 N/mm²

(ii) Total Extension of the Bar

Using the formula for bars of varying section:

ΔL = (P / E)   [ L₁/A₁ + L₂/A₂ + L₃/A₃ ]

Substituting the values:

ΔL = (35,000 / (2.1 × 10⁵)) × [ 200/(100π) + 250/(225π) + 220/(625π) ]

Calculate each term:

200/(100π) ≈ 0.6366
250/(225π) ≈ 0.3537
220/(625π) ≈ 0.1120

Sum = 0.6366 + 0.3537 + 0.1120 ≈ 1.1023

ΔL ≈ (35,000 / 210,000) × 1.1023 ≈ 0.183 mm

Final Answers:

• Stress in section 1 = 111.408 N/mm²
• Stress in section 2 = 49.5146 N/mm²
• Stress in section 3 = 17.8258 N/mm²
• Total extension of the bar ≈ 0.183 mm

Problem 1.09 – Bar with Reduced Middle Section

Question

The bar shown in Fig. 1.8 is subjected to a tensile load of 160 kN. The stress in the middle portion of the bar must not exceed 150 N/mm².

(a) Find the diameter of the middle portion.
(b) If the total elongation of the bar is required to be 0.2 mm, determine the length of the middle portion. Take Young’s modulus as E = 2.1 × 10⁵ N/mm².

Given

Tensile load, P = 160 kN = 160 × 10³ N
Permissible stress in middle portion, σ₂ = 150 N/mm²
Total elongation, ΔL = 0.2 mm
Total length of bar, L = 40 cm = 400 mm
Young’s modulus, E = 2.1 × 10⁵ N/mm²
Diameter of both end portions, D₁ = 6 cm = 60 mm

Area of each end portion:

A₁ = (π/4) D₁² = (π/4) × 60² = 900π mm²

Step 1 – Diameter of the Middle Portion

Let D₂ be the diameter of the middle portion, and A₂ its cross-sectional area.

σ₂ = P / A₂ ⇒ A₂ = P / σ₂ = 160000 / 150 ≈ 1066.7 mm²

Now,

A₂ = (π/4) D₂² ⇒ D₂² = 4A₂ / π ≈ 1358 mm² ⇒ D₂ ≈ 36.85 mm

Diameter of middle portion ≈ 36.9 mm.

Step 2 – Length of the Middle Portion

Let L₂ = length of the middle portion (in mm). Then the combined length of both large-diameter end portions:

L₁ = 400 − L₂

Total extension of the bar is the sum of extensions of the two end parts and the middle part:

ΔL = P/E × ( L₁/A₁ + L₂/A₂ )

Substituting values and ΔL = 0.2 mm, solving for L₂ gives:

L₂ ≈ 207.1 mm ≈ 20.7 cm

Final Answers

• Diameter of middle portion, D₂ ≈ 36.9 mm
• Length of middle portion, L₂ ≈ 20.7 cm

Problem 1.10 – Brass Bar with Axial Forces

Question

A brass bar of uniform cross-sectional area 1000 mm² is subjected to axial forces as shown in Fig. 1.9. Find the total change in length of the bar. Take E = 1.05 × 10⁵ N/mm².

Given

Area, A = 1000 mm²
Young’s modulus, E = 1.05 × 10⁵ N/mm²
Segment lengths: AB = 600 mm, BC = 1000 mm, CD = 1200 mm

Forces (after resolving the 80 kN at B into components):

• Segment AB: tensile load P₁ = 50 kN
• Segment BC: compressive load P₂ = 20 kN
• Segment BD: compressive load P₃ = 10 kN, acting over length BD = 2200 mm

Step 1 – Extension of AB

ΔL_AB = P₁ L₁ / (A E) = (50 × 10³ × 600) / (1000 × 1.05 × 10⁵) ≈ 0.2857 mm (increase)

Step 2 – Shortening of BC

ΔL_BC = − P₂ L₂ / (A E) = − (20 × 10³ × 1000) / (1000 × 1.05 × 10⁵) ≈ −0.1904 mm

Step 3 – Shortening of BD

ΔL_BD = − P₃ L₃ / (A E) = − (10 × 10³ × 2200) / (1000 × 1.05 × 10⁵) ≈ −0.2095 mm

Step 4 – Total Change in Length

Total ΔL = ΔL_AB + ΔL_BC + ΔL_BD = 0.2857 − 0.1904 − 0.2095 ≈ −0.1142 mm

Negative sign indicates an overall shortening of the bar.

Final Answer

Total change in length of the bar ≈ 0.114 mm decrease.

Problem 1.11 – Stepped Member with Point Loads

Question

A stepped member ABCD is subjected to point loads P₁, P₂, P₃ and P₄ as shown in Fig. 1.11. Given: P₁ = 45 kN, P₃ = 450 kN and P₄ = 130 kN.

(a) Determine the force P₂ required for equilibrium.
(b) Calculate the total change in length of the member. Assume E = 2.1 × 10⁵ N/mm².

Geometry and Properties

Segment AB: A₁ = 625 mm², L₁ = 120 cm = 1200 mm
Segment BC: A₂ = 2500 mm², L₂ = 60 cm = 600 mm
Segment CD: A₃ = 1250 mm², L₃ = 90 cm = 900 mm
Modulus, E = 2.1 × 10⁵ N/mm²

Step 1 – Equilibrium to Find P₂

Taking forces along the axis (right positive, left negative):

P₁ + P₃ = P₂ + P₄ ⇒ 45 + 450 = P₂ + 130 ⇒ P₂ = 365 kN

Step 2 – Internal Forces in Each Segment

From the equivalent system:

• AB is in tension by 45 kN
• BC is in compression by 320 kN
• CD is in tension by 130 kN

Step 3 – Extension/Shortening of Each Part

ΔL_AB = (45 × 10³ × 1200) / (625 × 2.1 × 10⁵) ≈ 0.4114 mm
ΔL_BC = − (320 × 10³ × 600) / (2500 × 2.1 × 10⁵) ≈ −0.3657 mm
ΔL_CD = (130 × 10³ × 900) / (1250 × 2.1 × 10⁵) ≈ 0.4457 mm

Step 4 – Total Change in Length

Total ΔL = 0.4114 − 0.3657 + 0.4457 ≈ 0.4914 mm

Final Answers

• Required force for equilibrium, P₂ ≈ 365 kN
• Total elongation of the member ≈ 0.491 mm (extension)

Problem 1.12 – Rod with Central Bore

Question

A tensile load of 40 kN acts on a steel rod of diameter 40 mm and length 4 m. A central coaxial bore of diameter 20 mm is to be drilled in the rod. To what length should the rod be bored so that the total extension increases by 30% under the same tensile load? Take E = 2 × 10⁵ N/mm².

Given

Load, P = 40 kN = 40,000 N
Original diameter, D = 40 mm
Bore diameter, d = 20 mm
Original length, L = 4 m = 4000 mm
Modulus, E = 2 × 10⁵ N/mm²

Original area of rod:

A = (π/4) D² = (π/4) × 40² = 400π mm²

Area removed by bore:

a = (π/4) d² = (π/4) × 20² = 100π mm²

Area of bored portion:

A − a = 300π mm²

Step 1 – Extension Before Boring

ΔL_before = P L / (A E)

Step 2 – Extension After Boring

Let the bored length be x metres = 1000x mm. Then unbored length = (4 − x) m = (4 − x) × 1000 mm.

ΔL_after = P/E × \left[ \frac{(4 - x) × 1000}{A} + \frac{x × 1000}{A - a} \right]

Required: ΔL_after = 1.3 × ΔL_before.

Solving this relation for x gives:

x ≈ 3.6 m

Final Answer

The rod should be bored over a length of approximately 3.6 m from the centre.

Problem 1.13 – Rigid Bar with Two Steel Wires

Question

A rigid bar ACDB is hinged at point A and supported in a horizontal position by two identical steel wires as shown in Fig. 1.12(c). A vertical load of 30 kN is applied at point B. Find the tensile forces T₁ and T₂ induced in the two wires.

Given

The two steel wires are identical, so they have the same cross-sectional area, the same modulus of elasticity, and the same original length. Thus:

A₁ = A₂, E₁ = E₂, L₁ = L₂

Distances from the hinge A (from figure):

• Wire at C (tension T₁) is at 1 m from A.
• Wire at D (tension T₂) is at 2 m from A.
• Load 30 kN at B is at 3 m from A.

Step 1 – Compatibility of Extensions

Because the bar is rigid and remains straight, the vertical deflections at C and D are related by similar triangles:

δ₁ / δ₂ = AC / AD = 1 / 2 ⇒ 2δ₁ = δ₂  ...(i)

But extensions of the wires are:

δ₁ = T₁ L / (A E) , δ₂ = T₂ L / (A E)

Substituting into (i):

2 T₁ = T₂  ...(ii)

Step 2 – Moment Equilibrium about A

T₁ × 1 + T₂ × 2 = 30 × 3  ...(iii)

Substitute T₂ = 2T₁ from (ii) into (iii):

T₁ × 1 + 2(2T₁) = 90 ⇒ 5T₁ = 90 ⇒ T₁ = 18 kN

Then from (ii):

T₂ = 2T₁ = 36 kN

Final Answers

• Tension in first wire at C, T₁ = 18 kN
• Tension in second wire at D, T₂ = 36 kN