Question

- how Attempt History Current Attempt in Progress X Your answer is incorrect. Aflanged-shaped flexural member is subjected to an internal axial force of P - 10.2 kN, an internal shear force of V - 14.6 kN, and an internal bending moment of M-2.1 kN-m, as shown. If the centraid is 29.29 mm above the bottom edge and the moment of inertia about the z axis is 511,283 mm, determine the normal stress oy at point K. 35 mm 6 min H V 15 mm M H 6 mm 65 mm А 15 mm 50 mm 6 mm 0.23.2 MPa O 29.7 MPa 027.6 MPa O OMP 24.7 MPa e Textbook and Media

Answer 1

NOTE:

ALL ANSWERS GIVEN ARE WRONG.

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35 mm 6 mm 15 mm H חומון 6 65 mm 15 mm 29.29 mm 50 mm 6 mm

The normal stress at K is comprised of:

• Direct Axial Stress
• Bending Stress

Direct Axial Stress =

P Oarial Area

where

• P = force = +10.2 kN = + 10200 N
• Area = Area of cross-section = $(35\times 6)+(50\times 6)+(6\times (65-6-6))=828\:mm^2$

P Oarial Area

$\sigma _{axial}=\frac{10200}{828}=12.318\:MPa$

Bending Stress =

Obending м ху І..

where

• M = Bending Moment = -2.1 kN.m = -2.1e6 N.mm [Compressive Stress will act]
• y = distance of point K from centroid = $29.29-15=14.29\:mm$
• I(zz) = Area moment of inertia = 511283 mm4

Obending м ху І..

$\sigma _{bending}=\frac{-2.1e6\times 14.29}{511283}=-58.694\:MPa$

TOTAL NORMAL STRESS AT K = -58.694 + 12.318 = -46.376 MPa

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As we can see, all the options given are wrong. But then, I have a strong feeling the answer is OPTION A.

I found that if we half the obtained answer, $\frac{-46.376}{2}=-23.188\:MPa\approx \mathbf{-23.2\:MPa}$

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I have checked my answer repeatedly. In case you happen to find any error that I have done, please feel free to ping in the comments. But other than that, I feel all the answers are wrong.

Kindly upvote in case I had/have helped you with the right answer. :)