Find the centroid (ˆˆ,xy) and Ix for the shape shown. Use a large rectangle and a small negative rectangle (which should have negative area and negative xI) and the origin a shown. If the shape in problem 5 is used for a beam with Mmax = 25 k-ft, what is the maximum bending stress? 10 in5) Find the centroid ( î, ŷ ) and Ix for the shape shown. Use a largerectangle and a small negative rectangle (which should have4 innegative area and negative I) and the origin a shown.6 in10 in6) If the shape in problem 5 is used for a beam with Mmax = 25 k-ft,what is the maximum bending stress?15.5 in5 inul t

Solution: Let large rectangle area A1=10×15.5=155in2 Let small rectangle area A2=6.5×6=39in2 We…

Answered: 10 in 5) Find the centroid ( â, ŷ ) and…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Find the centroid (ˆˆ,xy) and Ix for the shape shown. Use a large rectangle and a small negative rectangle (which should have negative area and negative xI) and the origin a shown.

If the shape in problem 5 is used for a beam with Mmax = 25 k-ft, what is the maximum bending stress?

10 in
5) Find the centroid ( î, ŷ ) and Ix for the shape shown. Use a large
rectangle and a small negative rectangle (which should have
4 in
negative area and negative I) and the origin a shown.
6 in
10 in
6) If the shape in problem 5 is used for a beam with Mmax = 25 k-ft,
what is the maximum bending stress?
15.5 in
5 in
ul t

Expert Solution

Step 1

Solution:

Let large rectangle area A₁=10×15.5=155in²

Let small rectangle area A₂=6.5×6=39in²

We need to find the position of centroid x' and y' from the x-axis and y-axis shown in above figure.

To find x-coordinate of the centroid

x-coordinate of centroid of area A₁ (x₁)= $\frac{10}{2} = 5 i n$

x-coordinate of centroid of area A₂ (x₂)= $4 + \frac{6}{2} = 7 i n$

Then position of x-coordinate of the centroid of the section is given by,

$\begin{array}{rcl} \hat{x} & = & \frac{A_{1} x_{1} - A_{2} x_{2}}{A_{1} - A_{2}} \\ = & \frac{155 \times 5 - 39 \times 7}{155 - 39} \\ = & 4.327 i n \end{array}$

To find y-coordinate of the centroid

y-coordinate of centroid of area A₁ (y₁)= $\frac{15.5}{2} = 7.75 i n$

y-coordinate of centroid of area A₂ (y₂)= $5 + \frac{6.5}{2} = 8.25 i n$

Then position of y-coordinate of the centroid of the section is given by,

$\begin{array}{rcl} \hat{y} & = & \frac{A_{1} y_{1} - A_{2} y_{2}}{A_{1} - A_{2}} \\ = & \frac{155 \times 7.75 - 39 \times 8.25}{155 - 39} \\ = & 7.581 i n \end{array}$

The position of centroid is (4.327,7.581)

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