All Exams Test series for 1 year @ ₹349 only
Question

The strength of a joint in a bolted connection is 620 kN. The size of plate is 180 mm width and 20 mm thick. If the yield strength of material of plate as 250 N/mm 2 and partial safety factor as 1.1, then the efficiency of the joint is ______.

The correct answer is

75.75%

Calculating Bolted Joint Efficiency

The efficiency of a bolted joint is determined by comparing the actual strength of the joint to the strength of the solid, unperforated plate. This ratio indicates how effectively the joint transfers the load compared to the original material.

We are given the following parameters for the bolted connection:

  • Strength of the joint ($P_{\text{joint}}$): 620 kN
  • Width of the plate ($W$): 180 mm
  • Thickness of the plate ($T$): 20 mm
  • Yield strength of the plate material ($f_y$): 250 N/mm$^2$
  • Partial safety factor for yielding ($\gamma_{m0}$): 1.1

Strength of Solid Plate

First, we need to calculate the design strength of the solid (gross) plate based on yielding. The gross area of the plate ($A_g$) is given by:

\( A_g = W \times T \)

Substituting the given values:

\( A_g = 180 \, \text{mm} \times 20 \, \text{mm} = 3600 \, \text{mm}^2 \)

The design strength of the solid plate against yielding ($P_d$) is calculated using the formula:

\( P_d = \frac{A_g \times f_y}{\gamma_{m0}} \)

Plugging in the values:

\( P_d = \frac{3600 \, \text{mm}^2 \times 250 \, \text{N/mm}^2}{1.1} \)

\( P_d = \frac{900000 \, \text{N}}{1.1} \)

\( P_d \approx 818181.82 \, \text{N} \)

To compare with the joint strength given in kN, we can convert this to kN:

\( P_d \approx \frac{818181.82}{1000} \, \text{kN} \approx 818.18 \, \text{kN} \)

Calculating Joint Efficiency

The efficiency of the joint (\( \eta \)) is defined as the ratio of the actual strength of the joint to the design strength of the solid plate, expressed as a percentage:

\( \eta = \left( \frac{\text{Strength of the joint}}{P_d} \right) \times 100\% \)

Substituting the given joint strength ($P_{\text{joint}} = 620$ kN) and the calculated solid plate strength ($P_d \approx 818.18$ kN or \(818181.82\) N):

Using values in N:

\( \eta = \left( \frac{620 \times 10^3 \, \text{N}}{818181.82 \, \text{N}} \right) \times 100\% \)

\( \eta \approx 0.7575 \times 100\% \)

\( \eta \approx 75.75\% \)

Alternatively, using values in kN:

\( \eta = \left( \frac{620 \, \text{kN}}{818.18 \, \text{kN}} \right) \times 100\% \)

\( \eta \approx 0.7575 \times 100\% \)

\( \eta \approx 75.75\% \)

The efficiency of the bolted joint is approximately 75.75%.

Was this answer helpful?

Important Questions from Riveted Connections

  1. The centre of adjacent rivets in the same row are separated by a distance known as: _____.

  2. What will be the value of bolt hole diameter for 12 mm diameter bolt as per IS 800-2007?

  3. If the working stress in bearing in power driven rivets is 300 N/mm2 and a double riveted double cover butt joint is used to connect plates of 12 mm thick, then the strength of rivet in bearing is

    (Given: The nominal diameter of rivet = 22 mm)

  4. Generally, the diameter of a rivet hole is made larger than the nominal diameter of the rivet by

  5. According to IS 800 : 1984, maximum axial tension in shop rivets in MPa is ______.

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App