Calculation of Stiffness in Structural elements

1)-AIM:-Compute lateral stiffness of the one story frame with an intermediate realistic stiffness of the beam. The system has 3 DOFs as shown. Assume L = 2h and El_b = El_c

ANS:-

• UNIT DISPLECMENT u1=1
• stiffeness coefficent of column transalation (2 similer)2*12EI/h^3
• stiffeness coefficent of beam roation (2 ends) EI/h^2
• UNIT DISPLECMENT u2=1
• stiffeness coefficent of column transalation at right end6EI/h^2
• stiffeness coefficent of column+ beam roation at left ends 4EI/h+ 4EI/L

4EI/h+ 4EI/2h=6EI/h

• • UNIT DISPLECMENT u3=1
  • stiffeness coefficent of column transalation at LEFT end6EI/h^2
  • stiffeness coefficent of column+ beam roation at RIGHT ends 4EI/h+ 4EI/L

  4EI/h+ 4EI/2h=6EI/h

fs=k.u

by applying statice condensation method we are deducating the 3 unknwn expression into2 unknown eq.

fs=(24Elc/h^3-Elc/h^36/7h(6h-6h)(1/1)u1=96ELc/7h^3u1

k=96Elc/7h^3

2)-aim-

For the following structures:

• Determine the number of degrees-of-freedom for dynamic analysis
• Establish the equation of motion
• Calculate their natural frequencies

ans-

• position of the element can be decribed by 1 coordinet only u(T)
• the system has single degree pf freedom str
• the column has fixed end at bottom .so the stiffnes will be as follow

k=3Ei/h^3

• the eq of motion shall be Mu+Ku=0
• natural frequncy formula w=k/m^1/2

b)aim-

ans-

• position of the element can be decribed by 1 coordinet only u(T) because beam with infinty stifness will include unifrom trnaslation in all column
• the system has single degree pf freedom str
• the column has fixed end at bottom.and 1 column has pinned -fixed ends the stifnss coefficent will be as follow
• so the stiffnes will be as follow

k1=12EI/h^3 k2=12EI/h^3 k3=3EI/h^3

for paraller stifness elemens Ke=(k1+k2+k3)=27EI/h^3

3)-aimConsider the propped cantilever shown in the figure below. The beams are made from the same steel section and have lengths as shown on the diagram. Determine the natural period of this system if a large mass, M, is placed at the intersection of the beams at point A. The weight of the beams in comparison with the mass M is very small.

ans-

• both elements have same material nad slection .it mens they have same moduls of elasticity and second momemt of ineria. but the length are differnt
• when a mas M IS APPLED AT the point both elements will respond simultensoly it mensthey are connected in parallel.
• effective stiffness of the system Ks=Kb+Kc
• Kb=48EI/lb^3
• Kc=3EI/lc^3
• Ks=3EI(1/lc^3+16/lb^3)
• natural frequncy formula w=k/m^1/2
• time peride T=2*3.14/W

4)-Determine an expression for the effective stiffness of the following systems:

A)-ANS:-

• THE STIFFENSS COEFFICENT OF THE BEAM ELEMENTS IS3 EI/L^2
• THE STIFFENSS COEFFICENT OF THE SPRING IS k
• THE STIFFENSS elements are in series so,

1/Ke=L^3/3EI+1/K

1/Ke=KL^3+3EI/3EIK

Ke=3EIK/KL^3+3EI

B-ANS:-

• THE STIFFENSS COEFFICENT OF THE BOTH COLUMN ELEMENTS IS 3EI/L^3
• THE STIFFENSS COEFFICENT OF THE SPRING IS k, BUT IS INCLINED .SO THE HORIZONTAL FORCE
• P AND DEFORMATION U HAS T BE RESOVELS INTO INCINES COMPONTS TO ARRIVE ACUTAL COEFFICENT OF HE SPRING
• P=p/cos
• u=ucos0
• k=p/ucos^2 0
• p/u=Ks-K COS ^0
• COS0=l/L^2+h62
• ks=kl*L^2/L^2=h^2
• ks=6EI/h62+KL/L^2+h^2

c)-ans-

1/ke=1/(48EI/L^3)+ 1/(4K* H^2/L^2+H^2)

Ke=(4AEh^2/L^2+h^2)48EI/L^3/(4AEh^2/L^2+h^2)48EI/L^3