A tie bar 5 mm × 8 mm is to carry a load of 80 kN. A specimen of the same quality steel of cross-sectional area 25 mm2 was tested in the laboratory. The maximum load carried by the specimen was 125 kN. Find the factor of safety in the design?
2.5
Calculating Factor of Safety for a Tie Bar
The question asks us to find the factor of safety for a tie bar based on its dimensions, the load it carries, and the material properties determined from a laboratory test on a specimen.
The factor of safety is a crucial concept in engineering design. It is defined as the ratio of the material's strength to the stress it experiences under the applied load. A higher factor of safety means the structure is designed to withstand loads significantly greater than the expected working load, providing a margin against failure.
Mathematically, the factor of safety (FOS) is often calculated as:
\( \text{FOS} = \frac{\text{Material Strength}}{\text{Working Stress}} \)
Let's break down the calculation:
1. Calculate the Working Stress in the Tie Bar:
The tie bar has a rectangular cross-section of 5 mm × 8 mm. The area of the tie bar is:
\( A_{\text{bar}} = 5 \, \text{mm} \times 8 \, \text{mm} = 40 \, \text{mm}^2 \)
The tie bar carries a load of 80 kN. We need to convert this load to Newtons:
\( P_{\text{bar}} = 80 \, \text{kN} = 80 \times 10^3 \, \text{N} = 80,000 \, \text{N} \)
The working stress (\( \sigma_w \)) in the tie bar is the load divided by the area:
\( \sigma_w = \frac{P_{\text{bar}}}{A_{\text{bar}}} = \frac{80,000 \, \text{N}}{40 \, \text{mm}^2} \)
\( \sigma_w = 2000 \, \text{N/mm}^2 \)
2. Determine the Material Strength from the Specimen Test:
A specimen of the same steel with a cross-sectional area of 25 mm\(^2\) was tested. The maximum load carried by the specimen was 125 kN.
\( A_{\text{specimen}} = 25 \, \text{mm}^2 \)
\( P_{\text{max}} = 125 \, \text{kN} = 125 \times 10^3 \, \text{N} = 125,000 \, \text{N} \)
The material strength (often taken as the ultimate tensile strength, \( \sigma_u \)) is the maximum load the specimen could carry divided by its original area:
\( \sigma_u = \frac{P_{\text{max}}}{A_{\text{specimen}}} = \frac{125,000 \, \text{N}}{25 \, \text{mm}^2} \)
\( \sigma_u = 5000 \, \text{N/mm}^2 \)
3. Calculate the Factor of Safety:
Now we can calculate the factor of safety by dividing the material strength by the working stress:
\( \text{FOS} = \frac{\sigma_u}{\sigma_w} = \frac{5000 \, \text{N/mm}^2}{2000 \, \text{N/mm}^2} \)
\( \text{FOS} = 2.5 \)
The factor of safety in the design is 2.5.
Let's summarize the values:
| Item | Value |
|---|---|
| Tie bar Area | 40 mm\(^2\) |
| Tie bar Load | 80,000 N |
| Working Stress (\( \sigma_w \)) | 2000 N/mm\(^2\) |
| Specimen Area | 25 mm\(^2\) |
| Maximum Specimen Load | 125,000 N |
| Material Strength (\( \sigma_u \)) | 5000 N/mm\(^2\) |
| Factor of Safety (FOS) | 2.5 |
The calculated factor of safety is 2.5, which means the material can theoretically withstand 2.5 times the working stress before reaching its ultimate strength.
Important Questions from Tension Member
A single angle in tension is connected by one leg only. If the areas of connecting and outstanding legs are respectively a and b, then what is the net effective area of the angle?
A) \(a-\frac{b}{1+0.35\times\frac{b}{a}}\)
B) \(a+\frac{b}{1+0.35\times\frac{b}{a}}\)
C) \(a-\frac{b}{1+0.20\times\frac{b}{a}}\)
D) \(a+\frac{b}{1+0.20\times\frac{b}{a}}\)
The net area of round bars to resist the tension, is the area of the cross-section at
The following are the statements about lug angle used to connect heavily loaded tension member to gusset plates.
(i) The length of end connection is reduced
(ii) By using lug angles there will be saving in the gusset plate
(iii) Cost of connection increases due to additional fasteners and angle required.
A steel plate is 300 mm wide and 10 mm thick. It has one rivet of nominal diameter 18 mm. The net sectional area of the plate is
The best tension member section will be as:
