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Grade 10Electric Current

each bar of thr plane truss is 1.6 m long and has cross sectional area 2.5 cm sq. and modulas of elasticity 2*10 ^6 kg/cmsq.the turnbuckle t in the bar ab is double acting with 13 threads per cm. how many turns n of the turnbuckle will be required to produce an axial tension of 5000 kg in the bar ab?

Profile image of Navjyot Kalra
12 Years agoGrade 10
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1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To determine the number of turns required for the turnbuckle to produce an axial tension of 5000 kg in the bar AB of the plane truss, we need to consider the relationship between the axial load, the properties of the material, and the mechanical advantage provided by the turnbuckle. Let's break this down step by step.

Understanding the Basics

The axial tension in a bar can be related to the elongation it experiences when a load is applied. The elongation (ΔL) can be calculated using the formula:

ΔL = (P * L) / (A * E)

Where:

  • P = axial load (in kg)
  • L = original length of the bar (in cm)
  • A = cross-sectional area (in cm²)
  • E = modulus of elasticity (in kg/cm²)

Given Data

  • Length of the bar (L) = 1.6 m = 160 cm
  • Cross-sectional area (A) = 2.5 cm²
  • Modulus of elasticity (E) = 2 × 10^6 kg/cm²
  • Axial tension (P) = 5000 kg
  • Threads per cm of the turnbuckle = 13

Calculating Elongation

Now, substituting the values into the elongation formula:

ΔL = (5000 kg * 160 cm) / (2.5 cm² * 2 × 10^6 kg/cm²)

Calculating the denominator:

2.5 cm² * 2 × 10^6 kg/cm² = 5 × 10^6 kg

Now substituting back into the elongation formula:

ΔL = (5000 * 160) / (5 × 10^6)

ΔL = 800000 / 5000000 = 0.16 cm

Determining the Turns of the Turnbuckle

The turnbuckle adjusts the length of the bar by a certain amount with each turn. The amount of adjustment per turn can be calculated using the pitch of the threads:

Pitch = 1 / (threads per cm) = 1 / 13 cm = 0.0769 cm per turn

Now, to find the number of turns (n) required to achieve the elongation of 0.16 cm:

n = ΔL / Pitch

Substituting the values:

n = 0.16 cm / 0.0769 cm ≈ 2.08 turns

Final Calculation

Since we cannot have a fraction of a turn in practical applications, we round up to the nearest whole number. Therefore, you would need approximately 3 turns of the turnbuckle to achieve the required axial tension of 5000 kg in the bar AB.

This calculation illustrates how mechanical systems can be analyzed using fundamental principles of mechanics and material properties. Understanding these relationships is crucial for designing and analyzing structural components effectively.