can you help me explain the problem above, its already finished, buy apparently

Question

can you help me explain the problem above, its already finished, buy apparently some steps are missing

1.13 Derive an expression for the equivalent stiffness of the system of Fig. 1-18 when the deflection of the machine is used as the generalized coordinate. E. I Fig. 1-18 Consider a concentrated downward load F applied to the midspan of the simply supported beam leading to a midspan deflection x. A compressive force kx is developed in the spring. The total downward force acting on the beam at its midspan is F-kx. As noted in Problem 1.8, the midspan deflection of a simply supported beam due to a concentrated load at its midspan is FL 48E1 Thus for the beam of Fig. 1-18, LJ 48E which leads to 48EI k+? The equivalent stiffness is obtained by setting F 1, leading to

Explanation / Answer

From the given diagram,
The mid-span deflection of a simply supported beam due to a concentrated load (F') at it's mid-span is
x=(FL^3)/(48EI) — (1)
Where F is the resultant force on the mid-span due to the interaction of the load due to mass 'm' which is F' with the spring force F"
F=F'-F"
Where F'=mg is the mass load
and F"=kx is the spring load
Therefore net force on the beam ,
F=F'-kx — (2)

From (1) and (2),
x=(F'-kx)(L^3/(48EI))

Now in order to find x,the term(L^3/(48EI)) is to be shifted from RHS to LHS
(48EI/L^3)x=(F'-kx)

Adding the term (kx) on both sides,
(48EI/L^3)x+kx=(F'-kx)+kx
=>(48EI/L^3)x+kx=F'
=>((48EI/L^3)+k)x=F'
=>((48EI/L^3)+k)=F'/x — (3)

Now the equivalent spring k(eq) is the ratio of net force on the system and the deflection
k (eq)=F'/x — (4)
{since the net spring force F'=k(eq)*x}

From (3) and (4),
F'/x=k (eq)=((48EI/L^3)+k)=(k+(48EI/L^3))

Therefore equivalent spring rate of the system=k (eq)=k+(48EI/L^3)

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