QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

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!this is the title of the program
program log1
implicit none 
!declaring the  variables
integer :: i,n
real :: sum,x
!Printing the statement to enter the value of x
PRINT *, 'Enter the value of X.'
!Reading the value of x from command line
READ *, x
!Printing the statement to enter the value of n
PRINT *,'Enter the value of n.'
!Reading the value of n from command line
READ *, n
!setting intial value of sum=x
sum=x
!using do loop to calculate our log(x+1) expension upto n terms
do i=2,n
sum=sum-((-x)**i)/i
print *, sum

!End do loop

end do

!end Program

end program log1