TEST BANK FOR Mechanics of Aircraft Structures 2'nd edition By C T sun (Solution Manual)

1.1 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 = − =
and the optimum ratio is = 1
a
b
and max 2
8
2 2 ( / 4) ( / 4) tL
T
L L t
T
abt
T =
⋅ ⋅ ⋅
τ = =
(b) Then corresponding max σ under the optimum condition stated above can
be obtained using M = T . We have
allowable allowable
allowable tL
T
t L L L
T
tb a b
M
σ τ
σ τ τ
2
2
3
2
12 3
( / 4) (3 / 4 / 4)
3
(3 )
3
max 2 max
< =
= = =
⋅ ⋅ ⋅ +
=
+
=
This means that when the structure fails in shear, the bending stress is
still within the allowable stress level. Thus the optimum ratio = 1
a
b
is
valid.
(4) In conclusion, = 1
a
b
achieves the most efficient section for the stated conditions.
--- ANS
1.1.3
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
1.2 Do problem 1.1 with M =αT where α = 0 to ∞.
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 = .
(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 M =αT 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
1.2.1
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
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 M =αT and b/a = 3 and check whether max τ is
within the allowable shear stress allowable τ .
allowable allowable
tL
M
L L t
M
abt
T
τ
α
σ
α
σ
α α
α
τ
1 2
1
3
32
2 ( / 8) (3 / 8)
/
2 max 2 max
= =
= =
⋅ ⋅ ⋅
= =
We have allowable allowable allowable τ τ
α
τ ≤τ ⇒ ≤ 2
max
􀃎 α ≥ 2 (since > 0 allowable τ is always satisfied)
(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 =αT .
(a)
abt
T
2 max τ =
Again we minimize S = 2abt = (L − 2b)bt in order to maximize τ , i.e.,
4
0 ( 4 ) 0
L
L b b
b
S = ⇒ − = ⇒ =
∂
∂
, so
2 4
L
b
L
a = − =
and the optimum ratio is = 1
a
b
and max 2
8
2 2 ( / 4) ( / 4) tL
T
L L t
T
abt
T =
⋅ ⋅ ⋅
τ = =
(b) Then corresponding max σ under the optimum condition stated above can
be obtained using M =αT . We have
allowable
allowable
allowable
tL
T
t L L L
T
tb a b
M
ασ
σ
ατ α
ατ
α α
σ
4
3
)
2
(
2
3
2
3
2
12 3
( / 4) (3 / 4 / 4)
3
(3 )
3
max 2 max
= = ⋅ =
= =
⋅ ⋅ ⋅ +
=
+
=
Since allowable allowable allowable σ ≤σ ⇒ ασ ≤σ
4
3
max
􀃎
3
4 α ≤ (since > 0 allowable σ is always satisfied)
1.2.2
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) From the above two approaches, we have the conclusions.
(i) For
3
4
0 <α ≤ , the failure is controlled by shear and the optimum ratio
of = 1
a
b
achieves the most efficient section..
(ii) For α ≥ 2, the failure is controlled by bending and the optimum ratio of
= 3
a
b
achieves the most efficient section.
(iii) For 2
3
4 <α < , the optimal ratio lies between 1 and 3. The most
straightforward way in finding the best ratio for a given α in this range
is to calculate the maximum bending moments and torques for different
values of b/a ratios between 1 and 3 and pick the ratio that produces the
greatest minimum failure load, either T or M.
--- ANS
1.2.3
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
1.3 The dimensions of a steel (300M) I-beam are b = 50 mm, t = 5 mm, and h = 200
mm (Fig. 1.17). Assume that t and h are to be fixed for an aluminum(7075-T6)
I-beam. Find the width b for the aluminum beam so that its bending stiffness EI
is equal to that of the steel beam. Compare the weights-per-unit length of these
two beams. Which is more efficient weightwise?
Figure