A 3ft long bar with a 1in x 1in cross section is subjected to a 14,153lb tensile load. What is the tensile stress in ksi? Answer to two decimal places. Type your answer...
A 3ft long bar with a 1in x 1in cross section is subjected to a 14,153lb tensile load. What is the tensile stress in ksi? Answer to two decimal places. Type your answer...
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Saeed Moaveni
Chapter7: Length And Length-related Variables In Engineering
Section: Chapter Questions
Problem 43P: A 10 cm long rectangular bar (when subjected to a tensile load) deforms by 0.1 mm. Calculate the...
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![**Problem Statement:**
A 3-ft long bar with a 1 in x 1 in cross-section is subjected to a 14,153 lb tensile load. What is the tensile stress in ksi? Answer to two decimal places.
*Type your answer...*
**Explanation:**
To find the tensile stress (σ) in ksi (kilo-pounds per square inch), use the formula:
\[ \sigma = \frac{F}{A} \]
Where:
- \( F \) = Force in pounds (lb)
- \( A \) = Cross-sectional area in square inches (in²)
Given:
- \( F = 14,153 \) lb
- Cross-sectional dimensions = 1 in x 1 in, so
- \( A = 1 \times 1 = 1 \) in²
Calculate the tensile stress in psi (pounds per square inch) first:
\[ \sigma = \frac{14,153 \, \text{lb}}{1 \, \text{in}^2} = 14,153 \, \text{psi} \]
To convert psi to ksi (1 ksi = 1,000 psi):
\[ \sigma = \frac{14,153}{1,000} = 14.15 \, \text{ksi} \]
Thus, the tensile stress is **14.15 ksi**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1815343-0e1e-45bc-b31a-0935809f9211%2Fbf0db897-b29a-434a-9991-dad046d01c11%2F1oukizh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 3-ft long bar with a 1 in x 1 in cross-section is subjected to a 14,153 lb tensile load. What is the tensile stress in ksi? Answer to two decimal places.
*Type your answer...*
**Explanation:**
To find the tensile stress (σ) in ksi (kilo-pounds per square inch), use the formula:
\[ \sigma = \frac{F}{A} \]
Where:
- \( F \) = Force in pounds (lb)
- \( A \) = Cross-sectional area in square inches (in²)
Given:
- \( F = 14,153 \) lb
- Cross-sectional dimensions = 1 in x 1 in, so
- \( A = 1 \times 1 = 1 \) in²
Calculate the tensile stress in psi (pounds per square inch) first:
\[ \sigma = \frac{14,153 \, \text{lb}}{1 \, \text{in}^2} = 14,153 \, \text{psi} \]
To convert psi to ksi (1 ksi = 1,000 psi):
\[ \sigma = \frac{14,153}{1,000} = 14.15 \, \text{ksi} \]
Thus, the tensile stress is **14.15 ksi**.
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