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 Elb = Elc
Solution :
if U1U1 = 1
U1U1 = 24 EIch3EIch3
'U2'′U2′= 6 EICh2EICh2
U3U3 = 6 EICh2EICh2
1 st equation :
If U2U2 = 1
U1=6EICh2U1=6EICh2
U2=4EIb2h+4EIChU2=4EIb2h+4EICh
U3=2EIb2hU3=2EIb2h
2nd equation :
If U3=1U3=1
U1=6EICh2U1=6EICh2
U3=4EIb2h+4EIChU3=4EIb2h+4EICh
U2=2EIb2hU2=2EIb2h
This is solved using a 3 X 3 matrix
*According to stiffness matrix method :
(U2U3)(U2U3) = - [6h2h2h26h2]−1[6h2h2h26h2]-1(6h6h)(6h6h)U1U1
−(136h4−h4)[−6h2h2h2−6h2]6h(11)U1-(136h4-h4)[-6h2h2h2-6h2]6h(11)U1
−6h35h4[6h2−h2−h26h2](11)U1-6h35h4[6h2-h2-h26h2](11)U1
−6h35h4(5h25h2)U1-6h35h4(5h25h2)U1
= −6h35h4×5h2(11)U1-6h35h4×5h2(11)U1=(U2U3)=(U2U3)
(U2U3)(U2U3)= −67h(11)-67h(11)U1U1`
substituting the above equation in the equation given below
fs=(24U1+6hU2+6hU3)EICh3fs=(24U1+6hU2+6hU3)EICh3
= fs=(24EICh3−67hEICh3[6h,6h](11))U1fs=(24EICh3-67hEICh3[6h,6h](11))U1`
=(24EICh3−67hEICh3.12h)U1(24EICh3-67hEICh3.12h)U1
=(EIch3(24−67h).12h)U1(EIch3(24-67h).12h)U1
=(EICh3(24)(−67h).12h)U1(EICh3(24)(-67h).12h)U1
= EICh3(24−727)U1EICh3(24-727)U1
=FS=EICh3967U1FS=EICh3967U1
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
A)
*The degree of freedom is 1
*Equation of motion
= mu + fd+fs= p(t)
= mu + ku = P(t)
= (12EIh3)U+Mu=P(t)(12EIh3)U+Mu=P(t)
* Natural frequency
= Fn=1TnFn=1Tn
= Tn=2πwnTn=2πwn
= Fn=wn2πFn=wn2π = √kMkM.12π.12π
=wn=√kmwn=km
= wn=√(12EIh3m)wn=(12EIh3m)
B)
*The degree of freedom is 1
*as the beam is rigid ( i.e flexural rigidity M.I = infinity
* K= ∑column12EIh3∑column12EIh3
= 36EIh336EIh3
*Equation of motion
= mu + fd+fs= p(t)
= mu + ku = P(t)
=mu + 36EIh336EIh3 u = p(t)
* Natural frequency
= Fn=1TnFn=1Tn
= Tn=2πwnTn=2πwn
= Fn=wn2πFn=wn2π = √kMkM.12π.12π
= wn=√kmwn=km
= wn=√36EImh3wn=36EImh3
3)
* Natural frequency
= Fn=1TnFn=1Tn
= Tn=2πwnTn=2πwn
= Fn=wn2πFn=wn2π = √kMkM.12π.12π
= wn=√kmwn=km
∴wn=√gu∴wn=gu
Tn=2π√ugTn=2πug
Tn=2π√ugTn=2πug
4)
a)
fsKe=fsks+fskbfsKe=fsks+fskb
Ke=KsKbKs+KbKe=KsKbKs+Kb
Kb=3ELL3Kb=3ELL3
Ke=K3EIL3K+3EIL3Ke=K3EIL3K+3EIL3
Ke=3EIK(3EI)+L3KKe=3EIK(3EI)+L3K
b)
*as the beam is rigid i.e EI = infinity
K= ∑column12EIh3∑column12EIh3
k ( column ) = 24EIh324EIh3
Ke=K+K(column)Ke=K+K(column)
Ke=K+24EIh3Ke=K+24EIh3
Ke=Kh3+24EIh3Ke=Kh3+24EIh3
C)
Ke=Ka+KbKe=Ka+Kb
Ke=EA√h2+L2Ke=EAh2+L2+48EIL3