Question

(1 - r)dr- (1/2)2

Answer 1

3.1.1 Consider the equations for the bar provided in the problem. du W=C d Here, w &u is the function of x. du W=C - fah ch Consider the variable s to avoid confusion. W= fds Consider the decreasing function f = 1- x, In terms of s, f =1-5 Thus, the expression for wis, w=S (1–s) ds WS- 2 =(1-1- 2 2 x2 1 W= 2 or *)- 1 1+- 2 2

Consider the above expression. du w(x)=C dx du w(x) dx с du dx C(x) Here, c is the constant. Consider the variable s to avoid confusion. S2 Thus, the function of w(s)= 2 2 w(s) du ds c(s) Here, c(s)= c is the constant function. =--S +

Substitute the respective values in the above expression. 1 1 du 을 S + ds с 2 2 Integrate the above expression with limits provided in the problem. s? 1 ſ du = SEC 2 -S + - 2 2 1 U= 2 S+ - ds 2 1 U=- CJO so s? --S + - 2 2 ds 1X 3 1 S S +- U= с 6 2 1 x x2 с 6 + - 2 2

3.1.2 For the hanging bar having weakening elasticity but the constant f, Consider the equations for the bar provided in the problem. du W=C ch Here, w&u is the function of x. Consider the variable s to avoid confusion. du fds ds Consider the constant function f. W=C = W= 1 w= f(5) w=f.(1-x) ОТ w(x) = f (1-x)

Consider the above expression. du w(x)=( dx du w(x) dx c(x) Consider the variable s to avoid confusion. = du ds C($) Substitute the respective values in the above expression. du f.(1-s) ds Here, c(s)=1-5 Integrate the above expression with limits provided in the problem. [ au= 51-5) Sa+s) (1-s)ds Si fos U= X U ds 0 u = f ($) u= f (x-0) u= f (x) u(x) = f(x)