Question 1) F=22 kN single load, M-24 kN.m moment and w=5 kN/m distributed load act on the beam whose loading condition is given in the figure. The length L is also given as L=4 m. The cross-sectional properties of the beam are; body Write your result in mm .) height y g width x f = 172 mm, flange height y f=16 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 B level of the beam. According to this; Question 1-E) Find the y distance of the section center with respect to the axis passing through the base of the section. ( = 203 mm , body thickness x g = 19 mm, flange Question 1-F) 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 .) Find the moment of inertia of the beam D abutment response at point (D y), Question 1-A) find. (Your result kN the size of your font.) ) Find the support response (A y) at point Question 1-B A. (Your result kN the size of your font.) Question 1-C ) Find the shear force (V B) at point B. ( Write your result in kN.) ciestion 1-D ) Find the bending moment (MBB) at point (Your result kn.m the size of your font.)

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=24 kN.m moment and
w=5 kN/m distributed load act on the beam whose loading
condition is given in the figure. The length L is also given as
L=4 m. The cross-sectional properties of the beam are; body
height y g = 203 mm , body thickness x g = 19 mm, flange
width x
Question 1-E) Find the y distance of the section center with
respect to the axis passing through the base of the section. (
Write your result in mm .)
Question 1-F)
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 .)
f
172 mm ,
flange height yf = 16 mm. Point E on
Find the moment of inertia of the beam
the section is located just below the flange-body
junction. It is desired to determine the stress state in the
section taken from the B level of the beam. According to this;
D abutment response at point (D y
Question 1-A)
find. (Your result kN the size of your font.)
) Find the support response (A y ) at point
Question1-B
A. (Your result kN the size of your font.)
Question 1-C ) Find the shear force (V B) at point B. ( Write
your result in kN.)
В.
luestion1-D ) Find the bending moment (MB B ) at point
(Your result kn.m the size of your font.)
Transcribed Image Text:Question 1) F=22 kN single load, M=24 kN.m moment and w=5 kN/m distributed load act on the beam whose loading condition is given in the figure. The length L is also given as L=4 m. The cross-sectional properties of the beam are; body height y g = 203 mm , body thickness x g = 19 mm, flange width x Question 1-E) Find the y distance of the section center with respect to the axis passing through the base of the section. ( Write your result in mm .) Question 1-F) 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 .) f 172 mm , flange height yf = 16 mm. Point E on Find the moment of inertia of the beam the section is located just below the flange-body junction. It is desired to determine the stress state in the section taken from the B level of the beam. According to this; D abutment response at point (D y Question 1-A) find. (Your result kN the size of your font.) ) Find the support response (A y ) at point Question1-B A. (Your result kN the size of your font.) Question 1-C ) Find the shear force (V B) at point B. ( Write your result in kN.) В. luestion1-D ) Find the bending moment (MB B ) at point (Your result kn.m the size of your font.)
w kN/m
F kN
M kN.m
A
В
C
2L
L
Xf
Ti
Yf
E
Yg
Xg
Transcribed Image Text:w kN/m F kN M kN.m A В C 2L L Xf Ti Yf E Yg Xg
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