**Structural Component Analysis**A structural component in the form of a flat plate, with a thickness of 20.8 mm, is constructed from metal with the following properties:- Yield strength: 533 MPa- Critical fracture toughness: 22.0 MPa-m\(^{1/2}\)**Problem:**Assume that a crack develops in the plate. Given that the geometry of the crack results in a \( Y \) value of 1.5, determine the critical crack length if the design stress is 0.3 times the yield strength.**Equations:**1. **Stress Concentration Factor**: \[ K_t = \frac{\sigma_m}{\sigma_0} = 2 \left( \frac{a}{\rho_t} \right)^{1/2} \]2. **Fracture Toughness**: \[ K_c = Y \sigma_c \sqrt{\pi a} \]**Explanation:**- \( \sigma_m \) is the maximum stress.- \( \sigma_0 \) is the nominal stress.- \( a \) is the crack length.- \( \rho_t \) is the radius of the notch tip; not specified but may affect stress concentration.- \( K_c \) is the critical stress intensity factor.- \( Y \) is a geometric factor for the crack.- \( \sigma_c \) is the critical stress (which is 0.3 times the yield strength in this problem).To solve for the critical crack length, substitute the appropriate values and solve the equation for \( a \). This analysis helps in understanding the parameters affecting material failure, aiding in safe and efficient design.

Given, Thickness of the flat plate = 20.8 mm Yield strength =σy=533 MPa Design stress= σc=0.3×σy=0.3×533=159.9 MPa Critical fracture toughness =Kc = 22 MPa-m1/2 Y=1.5From the equation: Kc=Yσcπa 22=1.5×159.9×πa⇒a=1π221.5×159.92=0.002678 m=2.678 mm Answer:"The critical crack length is 2.678 mm".

Engineering

Civil Engineering

A structural component in the shape of a flat plate 20.8 mm thick is to be made from a metal with yield strength of 533 MPa and a critical fracture toughness of 22.0 MPa-m!2. Assume a crack forms in the plate, and the geometry of the crack yields a Y value of 1.5. If the plate is designed to a design stress equivalent to 0.3 times the yield strength, what is the critical crack length? Equations: 1/2 а Kț = Om %3D 2(: Ke = YocTa па Pt

A structural component in the shape of a flat plate 20.8 mm thick is to be made from a metal with yield strength of 533 MPa and a critical fracture toughness of 22.0 MPa-m!2. Assume a crack forms in the plate, and the geometry of the crack yields a Y value of 1.5. If the plate is designed to a design stress equivalent to 0.3 times the yield strength, what is the critical crack length? Equations: 1/2 а Kț = Om %3D 2(: Ke = YocTa па Pt

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Crack Detection

Concrete materials are meant for supporting structures, building foundations and they act as several load-bearing structures. Over time, due to environmental conditions, and based on the constituent materials used, concrete materials are subjected to a great degree of distress. Corrosion of the reinforcement bars due to chemical influences, uneven base, earthquake, drying shrinkages, and alkali-aggregate reactions are some of the reasons for concrete distress.

Question

**Structural Component Analysis**

A structural component in the form of a flat plate, with a thickness of 20.8 mm, is constructed from metal with the following properties:
- Yield strength: 533 MPa
- Critical fracture toughness: 22.0 MPa-m\(^{1/2}\)

**Problem:**
Assume that a crack develops in the plate. Given that the geometry of the crack results in a \( Y \) value of 1.5, determine the critical crack length if the design stress is 0.3 times the yield strength.

**Equations:**

1. **Stress Concentration Factor**:
\[
K_t = \frac{\sigma_m}{\sigma_0} = 2 \left( \frac{a}{\rho_t} \right)^{1/2}
\]

2. **Fracture Toughness**:
\[
K_c = Y \sigma_c \sqrt{\pi a}
\]

**Explanation:**

- \( \sigma_m \) is the maximum stress.
- \( \sigma_0 \) is the nominal stress.
- \( a \) is the crack length.
- \( \rho_t \) is the radius of the notch tip; not specified but may affect stress concentration.
- \( K_c \) is the critical stress intensity factor.
- \( Y \) is a geometric factor for the crack.
- \( \sigma_c \) is the critical stress (which is 0.3 times the yield strength in this problem).

To solve for the critical crack length, substitute the appropriate values and solve the equation for \( a \). This analysis helps in understanding the parameters affecting material failure, aiding in safe and efficient design.

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