F=22 kN single load, M=26 kN.m moment and w=17 kN/m distributed load are acting on the beam whose loading condition is given in the figure. The length L is also given as L=5 m. The cross-sectional properties of the beam are; body height yg=239 mm, body thickness xg=17 mm, flange width xf=185 mm, flange height yf=15 mm. Point E on the section is located just below the flange-body junction. It is desired to determine the stress state in the section taken from the C level of the beam. According to this; Question 1-D) Find the moment of inertia of the beam section. (Do your operations by converting the lengths to meters. Take at least 4 digits for the decimal part of your result and use the expression E for the decimal part. Write it as 5E-3 instead of 0.005.) Question 1-E) For a section taken from the C point of the beam; Find the normal stress at point E on the beam section due to the bending moment. (Write your result in MPa.)

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Question 1) F=22 kN single load, M=26 kN.m moment and w=17 kN/m distributed load are acting on the beam whose loading condition is given in the figure. The length L is also given as L=5 m. The cross-sectional properties of the beam are; body height yg=239 mm, body thickness xg=17 mm, flange width xf=185 mm, flange height yf=15 mm. Point E on the section is located just below the flange-body junction. It is desired to determine the stress state in the section taken from the C level of the beam. According to this;

Question 1-D) Find the moment of inertia of the beam section. (Do your operations by converting the lengths to meters. Take at least 4 digits for the decimal part of your result and use the expression E for the decimal part. Write it as 5E-3 instead of 0.005.)

Question 1-E) For a section taken from the C point of the beam; Find the normal stress at point E on the beam section due to the bending moment. (Write your result in MPa.)

w kN/m
M kN.m
В
A
3
C
4/5
-
2L
2L-
F kN
Xf-
Yf
E
Yg
Xg
Yf
Xf
Transcribed Image Text:w kN/m M kN.m В A 3 C 4/5 - 2L 2L- F kN Xf- Yf E Yg Xg Yf Xf
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