ENGINEERING TRIPOS PART IB JUNE 2012
CHAIRMAN PROF. P DAVIDSON
•
k
MONDAY 4 TH JUNE 2012
m
2 TO 4
=
IX
PAPER 2 SOLUTIONS - STRUCTURES
AUTHORS:
DR. F CIRAK
DR. D MCCONNELL
Created by Diane Unwin
Friday, July 13,2012
Part IB, Paper 2 - Structures, 2011/12
1.
(a)
L
The second moment of area for the thin-walled square is
-2L3 t
3
The second moment of area for the cruciform is
The ratio of the second moments of both cross-sections has to be equal
l~
4
[e
xx
viz
I
FC
Structures
PartIB
(b)
L
W
The torsion constant for a thin-walled closed cross-section is defined as
4A2
J= __
e
J: ds
dt
.r
The torsion constant of the square is
The torsion constant of the equilateral triangle is
I
J3
J3
2
Ae=-W-W -W
2
2
4
4
=;,-f _ 3W t _ ~ 3
- 4-35 - 4 W t
The ration of the torsion constants is
(c)
2
Structures
PartIB
FC
x
Parameters for one (horizontal) semi-circle
Centroid
L
17:-t11s
=
2
!o1r: (L- sin 0 )
02
2
2
Lt dO = - L t cos 0 11r: = Lt
2
4
02
L
=> 11s =
17:
Second moment ofs area
I~~
~~
=
=> I~c; =
=>
1r: 112-dO
Lt
(L
= !o1r:
-sinO
2
0
2
!o0
)2 -dO
Lt
2
L3 t 17:
16
r _ L3 t 17:
16
11"1 -
Second moments of area around the centroid
1';1 =
L3 t 17:
16
Second moment of area for the entire cross-section
lu
= 2 (~'"
+"~ (~+~;) ') +21';'
L3t 17:
2(
16
L3 t 17: L 3t
= -- - 2
17:
L3t
217:
+
17:L3t
8
+2+
L3 t
+ L 3t + -
17:
L 3t
L 3t
217:
L 3t17:)
+16
= 2.5708L 3t
Alternatively, the second moments of area for the semi-circle can be determined with
the equations for the curved rod given in Mechanics Data Book
e _ 17:Ltk2
I rl'l1' - 2 x
3
Part IB
Structures FC
-------------------------------------------------------------:rr:
with a = ~ and a
2"
Torsion constant of the entire cross-section
Ae
= L 2 + 2rc (L)2
2" = L2 + rcL2
f
2.
dS
2
= 2rcL
t
dt
(a) Maximum displacement using virtual work
w
r
,'1--------
..
I
x
Virtual moment distribution M(x)
=
I·x Moment distribution due to w I
I
I w
IX3
M(x) -xw(x)-x = -~-x = --w
2
3
6 L
6L Principle of virtual work r
I wL
EI 6L Jo x dx = EI 30
I w
I· <5
L 4
4
(b) i. Although the system is statically indeterminate the bending moments in the beam
can be determined without making use of the force method.
Vertical deflection at Point A
4
Structures PartIB
I . SvEI
=
J
FC
M(x)M(x) dx
= ~ wIL2 L
8
8
wt L4
2
+
(2
L
6
3WIL4
2
2
w IL
WIL2) ~ _ ~~ w IL
8 + 32
2 32 8
4
wt L4
wIL
128 + 128 48 96
Alternatively, the vertical deflection can be obtained by superposing Data Book
cases.
Rotation of the left support
wt L3
24EI -
L2 L
Wt 32 6EI
w I L2 L
-8- 3EI
=
L3
-WI 192EI
Vertical deflection
Sv
=
L4
L3 L
wl128EI +wI192E12
ii. Horizontal deflection at Point A
To determine the horizontal deflection, first the moments in the columns need to be
determined. To this end, consider a statically determinate system where the joint
between the left column and beam is released
Compatibility
alX+ao=O
Horizontal displacement due to virtual unit load
2L3
al = 3EI
Horizontal displacement due to W2
W2 L 4
30EI
5
FC
Structures
Part IB
1
L
L
Horizontal force at the joint
After knowing the shear force at the top of the left column its horizontal deflection
can be computed
4
2
= I ~ L W2 L = 1 w2L h
EI3
20
EI 60 Note, this deflection is the same as for a single cantilever beam with twice the EI. a
3.
(a)
Axial stress pR
2t
O'a
250· 103 . I 41.667MPa
2.3.10- 3 Hoop stress pR
t
O'h = -
Shear stress = 2 -41.667 MPa = 83.33MPa
600.103 T
r---- 21CR2 t - 2ll· I -3· 10- 3
(b)
6
31.83MPa
Fe
Structures
Part IB
(c)
Von Mises criterion
This gives for plain stress
(0"3
= 0)
From the Mohr's circle we obtain
0"1
=
~()"
+
4
7
J+
r2
_1 ()"2
16
FC
Structures
PartIB
o3r-____~~____~______~~~--~------+_-~
The square root expression in the previous equations is abbreviated with R.
~a2 + ~ aR + R2 + ~ a 2 - ~aR + R2 _ ~ a 2 + R2
16
2
16
2
16 y2
9 a2 +3R2 = y2
16 ~a2+3't'2+2.a2 = y2
16
16 ~a2+3't'2
4
8
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~
p
r
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