Please answer questions 6,7,8,9,10 QUESTION If the end fixsity constant K-1 and
ID: 2086430 • Letter: P
Question
Please answer questions 6,7,8,9,10 QUESTION If the end fixsity constant K-1 and the length of column is 25 inches, the effective length of the (La) (how many 2) inches QUESTION 7 When we determine the allowable load on a colume wthout bucking,we are requred to use the practical value of the colu O True en fity constant K True or false False QUESTION the end foty constat K= 1 and the length of column is 25 nches This column is. O Fixed at both nd O Fixed at one end and free at e otheriend O Fixed on one end and pinned on the other end O Pinned at both ends QUESTIONS For a colu thit 30 inche·beseer, two supports the teoreta' cour n and f ty con unt K?50 and the prae beal column end taty eonsnt k as what the asme eksm t in whose nches? O 30 O 195 155 None of the above QUESTION 10 when calculating the slenderness ratio for a cylindncal column whose dameter has already been selected frsne least vale of radus of graion must be determ red which one of the blowng wil gve the least value of the radus of gyrahion? Smaller value of cross sectional area O Smaller value of Moment of Inertia Larger value of moment of inertia O None of the aboveExplanation / Answer
6) Answer 25 Inches
Effective Lenth of the Column = Actual Length if the Column / ? And Fixity Coefficient
Given the Actual Length of Column L = 25 Inch
And Fixity Coefficient K=1
Effective Lenth of the Column Le = L/?K = 25/?1 = 25 Inches
Effective Length of the Column is defined as distance between successive innflection points of Zeromoment points
7) False
Euler has developed a Formula to find the Critical Load on A column to Cause the Bukling Load. The Critical Load is the maximum axial Load a column can support when it is on verge of buckling.
Eulers Critical Load Pcr = ?2 EI / (Le)2
where Le = Effective Length = L/?K and K =Theoritical And Fixity Coefficient
Pcr = K?2 EI / L2
So to Determine the Safe Design load on Column we use the Theoritical Value of And Fixity Coefficient(K) but not Practical Value.
8) Pinned(Hinged) End columns
Based on the Vlaue of K the columns can be identified as below
9) Answer is None of the above
Effective Lenth of the Column = Actual Length if the Column / ? And Fixity Coefficient
Le = 30/?0.5= 42.42 Inch if we use Theoritical value
Le = 30/?0.65 = 37.21inch If we use Practical Value but none of them are given in options
10) Smaller Value of Moment of Inertia
I = Ak2 where K = Radius of Gyraiton
K = ?I/A In order to Decrease K the a) Moment of Inertial Should be Less or b) Area of Column should be More
SO correct answer is Smaller Value of Moment of Inertia
End Conditions And Fixity Coefficient Effective Length Both Ends Fixed 4 0.5L One End Fixed and One End Hinged/Pinned 2 0.707L Both Ends Hinged/Pinned 1 L One End Fixed - One End Free 0.25 2L