TEST BANK FOR Mechanics of Aircraft Structures 2\'nd edition By C T sun

The beam of a rectangular thin-walled section (i.e., t is very small) is designed
to carry both bending moment M and torque T. If the total wall contour length
(see Fig. 1.16) is fixed, find the optimum b/a ratio to achieve the
most efficient section if
L = 2(a + b)
M = T and allowable allowable σ = 2τ . Note that for closed
thin-walled sections such as the one in Fig.1.16, the shear stress due to torsion is
abt
T
2
τ =
Figure 1.16 Closed thin-walled section
Solution:
(1) The bending stress of beams is
I
My σ = , where y is the distance from the neutral
axis. The moment of inertia I of the cross-section can be calculated by considering
the four segments of thin walls and using the formula for a rectangular section
with height h and width w. wh Ad )
12
1
I = Σ( 3 + 2 in which A is the
cross-sectional area of the segment and d is the distance of the centroid of the
segment to the neutral axis. Note that the Parallel Axis Theorem is applied. The
result is (3 )
6
) ]
2
( ) (
12
1
2 [
12
1
2
2
3 3 2 a b
b tb
I = ⋅ tb + ⋅ ⋅ at + at ⋅ ≈ + , assuming that t is
very small.
(2) The shear stress due to torsion for a closed thin-walled section shown above is
abt
T
2
τ = .
1.1.1
Name: Mohamed Naleer Abdul Gaffor Email: [email protected] IP: 184.162.144.24
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
Mechanics of Aircraft structures
C.T. Sun
(3) Two approaches are employed to find the solution.
(i) Assume that the bending stress reaches the allowable allowable σ first and find
the corresponding bending maximum bending moment. Then apply the stated
loading condition of T = M to check whether the corresponding max τ has
exceeded the allowable shear stress allowable τ . If this condition is violated, then
the optimized b/a ratio is not valid.
(a)
(3 )
3
(3 )
6
| 2 2
2 tb a b
M
a b
tb
b
M
I
My
b
y +
=
+
⋅
= =
=
σ
When given L = 2(a + b) as a constant, a can be expressed in terms of b
and L as b
L
a = −
2
. Then we can minimize
6
(3 4 )
3
tb(3a b) tb L b
S
−
=
+
= in order to maximize σ , i.e.,
8
3
(3 8 ) 0
6
0
L
L b b
t
b
S = ⇒ − = ⇒ =
∂
∂
, so
2 8
L
b
L
a = − =
where the optimum ratio is = 3
a
b
Thus, max 2 3
32
(3 / 8) (3 / 8 3 / 8)
3
(3 )
3
tL
M
t L L L
M
tb a b
M =
⋅ ⋅ ⋅ +
=
+
σ =
(b) Check max τ with T = M and b/a = 3 and check whether max τ is within
the allowable shear stress allowable τ .
2
3
32
2 2 ( / 8) (3 / 8) max 2 max
allowable
allowable
allowable tL
M
L L t
M
abt
T
σ
τ
τ σ σ
> =
= = =
⋅ ⋅ ⋅
= =
The result above means that under this assumption, shear stress τ would
reach the allowable stress allowable τ before σ reaches allowable σ . Consequently,
the optimal ratio obtained is not valid and different assumption needs to be
made.
(ii) Assume now that failure is controlled by shear stress. We assume that
allowable τ =τ max is reached first and then find the corresponding bending stress
according to the loading condition M = T .
(a)
abt
T
2
τ =
Again we minimize S = 2abt = (L − 2b)bt in order to maximize τ , i.e.,
1.1.2
Name: Mohamed Naleer Abdul Gaffor Email: [email protected] IP: 184.162.144.24
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN
Mechanics of Aircraft structures
C.T. Sun
4
0 ( 4 ) 0
L
L b b
b
S = ⇒ − = ⇒ =
∂
∂
, so
2 4
L
b
L
a = −