ME2114 – Pointers Pointers to Worksheet on Energy Methods
/0
Q1. Convert the indeterminate problem to a determinate problem with unknown reaction force, R. Find R from .
⟹ Truss analysis:
D
R
0
∑ 2 ⟹ , 56 ,, , + 23 ,, /0 ⁄
Q2. Convert the indeterminate problem to a determinate problem with unknown reaction force, R. Find R from . Deflection of left end is given by .
P
L
A
B 3
k= EI/L
x P
⇓ L
A
B
3
k= EI/L
0 ( )) 1 + 2 + + 2 R
1
ME2114 – Pointers to Worksheet on Energy Methods
|=
Q3. To find the deflection at B, add a force Q at B and apply Castigliano’s Theorem,
A
|=
P
B Q
A common difficulty is to obtain expressions of T and M in the curved beam. From the free body diagram below and the definition of used,
, ≤≤0 2 {+, 0≤≤ 2 1+ , ≤≤0 2 {1+1+, 0≤≤ 2
T
M
A B
P Q=0
− 2 + 2 + 2 + 2 = Hint : Do the differentiation first, then set Q=0 before attempting the integration. You will find that the integral from – /2 to 0 becomes zero and the integral from 0 to /2 is greatly simplified.
2
ME2114 – Pointers to Worksheet on Energy Methods
Q4. Convert the indeterminate problem to a determinate one by looking at half the ring carrying half the load Q=P/2 as shown. Note that there will be a bending moment M 0 when the ring is ‘cut’ in half since the two halves are from a single continuous ring, i.e., M0=0 if the 2 halves are hinged together. This bending moment keeps the top and bottom ends of the half-ring from rotating, i.e.,
0
Q=P/2
P/2
M0 A
M M 0 QR sin
U
M 2
2 EI Rd 0
P/2
Determine M 0 in terms of Q from
0
. Then find
=/
Q5. Convert the indeterminate problem to a determinate one by replacing the support at C with a vertical reaction force Q and a reaction moment M c. P A
P
B
A x
R L
B
C
R
C Mc Q
Use Castigliano’s Theorem to find Q and M C by solving simultaneously,
For section BC: For section AB:
0, 0 / 1 + , ∫ ++ +, ∫ 3
ME2114 – Pointers to Worksheet on Energy Methods
Q6. Before applying PMPE, enforce y=0 at x=0, and y=0 at x=L.
Q7. y2 should give a better approximation because it has zero s lope at the ends to correctly represent fixed boundary conditions.
Π+ 2 + + , + +
Q8. Find the PE in terms of u1 , u2 and u3 from
, where
Q9. The following are intermediate results for you to check. Rod
k
C
AC
EA / 4 3
6 kN
AB
EA / 4
-30o
3/2
-1/2
BC
EA / 4
30o
3/2
1/2
4 3m
30
A 4m
o
30 B
o
4m
cos
sin
0o
1
0
EA=50e6 N
0 0.75 0.4330 u B 0 1.5 0 0.5 0.4330 0.25 0 B y EA (1) 0.4330 uC 0 0.4330 1.3274 4 0.75 0.4330 0.25 vC 6e3 0.4330 0.25
0.4330 u B 0 u B 0.0011 0 u 0.0014 (2) 0.75 1.3274 0.4330 uC C 4 0.4330 0.4330 vC 0.0063 0.25 vC 6e3
1.5
EA
0.75
and from the deleted row in (1), B y
EA
4
0.4330uc 0.25vc 12000 N
4
ME2114 – Pointers to Worksheet on Energy Methods
Q10. The following are intermediate results for you to check.
D
C
2m A B P 2m
2m
Rod
k
AC
EA / 2
AB
cos
sin
45o
1/
1/
EA / 4
0o
1
0
CD
EA / 2
180o
-1
0
BC
EA / 2
135o
-1 /
2
2
2
2
1/
2
2
EA=80e6N
1.2071 0.5 0.5 EA 0.5 0.5 2 2 0.5 0.5 0.5
0.5 0.5 2.4142 0
u B 0 0.5 10e 3 B y uC 0 0 vC 0 1 0.5
The 1st, 3rd and 4th eqn of the above simultaneous eqns can be written as
0 EA (0.5)(10e 3) 0.5 u B 0.5 1.2071 0 2 2 0.5 0.5 B y 0.5 0.5 10e 3 EA EA (0.5)(10e 3) 0 2.4142 0 uC 0 2 2 0.5 2 2 EA 0 0 1 vC e 0 ( 0 . 5 )( 10 3 ) 0.5 2 2 u B 0.0017 uC 0.0017 vC 0.0041
5
ME2114 – Pointers to Worksheet on Energy Methods
Q11. You only need to consider half the problem because of symmetry and you only need the 2nd equation in [ K ][u]=[f ] to solve for va. A
Zero x displacement and zero y reaction force A
√ 2
y
D
√ 2 ⇒ 4 7 8 00 0 +(⁄2√ 2)8(⁄(√ ⁄22√ )02)+7(⁄√ 2) ⁄2√ 0 2 ⁄⁄√ 2√ 22 [ ⁄√ 2 ] []
C L/2
fixed
L/2
B
x
fixed
(**) Solve va from 2nd eqn in [K][u]=[f] i.e.,
So only need to determine 2 nd row of [K] matrix
Q12.
y ~ Using the given formula B 0 x for the 2 elements below,
~
B
0
y
0
y
x
0
x
0
y
x
y
x
x , it can be shown that y 0
~
a
B
b
Element a
Element b 2L
L
L 2L
This means
k
local b
because k
k
local a
u P 1 v 2 1
For given problem, [K ]
c
c
local
t
B~ EB~ . T
4 A
Et 0.6563 2
L
6
0
u 1.2188 v 0
c
c
1 2 1
P
