GATE Aerospace Coaching by Team IGC Aircraft Structures Basics
Transcript of GATE Aerospace Coaching by Team IGC Aircraft Structures Basics
GATE Aerospace Coaching by Team IGC
Aircraft Structures Basics
Bredth Betho Theory
Torsion of thin walled section
Assumption:-
Wapping displacement are freely permitted
ππ₯π is presents and every other stress component is zero
ππ₯π doesnβt vary along thickness direction
Force equilibrium along Z-axis
βπ1π‘1ππ§ + π2π‘2ππ§ = 0
π1π‘1 = π2π‘2 = π
Shear flow per unit length
q in terms of torque
Diagram
ππ = πππ . π
π = β« ππ = β« ππ ππ = 2 β« π1
2π ππ
π = 2 β« π ππ΄
π = 2π β« ππ΄
π = 2π΄π
π =π
2π΄
Angle of Twist
Shear strain energy stored in the structure
=1
2ππβ² β πππππ ππ π‘π€ππ π‘
= β«1
2. ππ£ . π£πππ’ππ
= β«1
2
π2
πΊ(π‘ππ π 1)
1
2ππβ² = β«
π2
2πΊπ‘ππ
πβ² = β«π2
ππΊππ
= β«π2
π‘ππΊππ
πβ² = β«π
2πππΊππ
Torsional Rigidity
πΊπ =π
πβ² ; π½ =
4π΄2
β«ππ
π‘
πΊπ =4π΄2
β«ππ
πΊπ‘
β π‘πππ π‘ππππ π ππππππ‘π¦
π½ = πΌπ β πππ πππππ’πππ ππππ π πππ‘πππ
Problems:-
(1). In a thin walled rectangular subjected to equal and opposite forces as shown in fig, the shar
stress along lay is.
(a). Zero (b). constant non-zero (c). yasles linearly (d). yaslesparabolitally
Sol:-
From breathbetha theory
π = 2π΄π
π =π1
2+
π2
2
π =π
2π΄
π = π2 ππ‘ =π
2π΄
(2). A thin walled tube of circles cross section with mean radius R has a central way which
divide it into two symmetrical cell a torque on is acting on the section. The shear flow q in
central web is
(a). π =π
2ππ2 (b). π = 0 (c). π =π
4ππ2 (d). π =πΎ
ππ2
Sol:-
π = π1 β π2
π = 2π΄1π1 + 2π΄2π2
π1β² = π2
β² (ππππ ππππππ‘ππππππ‘π¦ πππ’ππ‘πππ)
π1β² = β«
π1ππ
2π΄1πΊπ‘1
π2β² = β«
π2ππ
2π΄2πΊπ‘2
π1β² = π2
β²
β«π1ππ
2π΄1πΊπ‘1= β«
π2ππ
2π΄2πΊπ‘2
π1 = π2
π = 0
(3). An Euler Bernoulliβs Beam has rectangular cross section Beam shear in fig and subjected to
non-uniform BM along its length π£2 =ππ
π£2π¦, the shear stress distribution ππ₯π₯ across the cross
section
(a). ππ₯π₯ =ππ
2πΌπ¦(β
2) (b). ππ₯π₯ =
ππ(β
2)
2
2πΌπ¦(1 β
π2
β
2
) (c). ππ₯π₯ =ππ
2πΌπ¦(
2β
2
)
2
(d). ππ₯π₯ =
π2(β
2)
2
2πΌπ¦
(4). Find the torsional constant for a ring of radius is
π½ =4π΄2
β«ππ
π‘
=4(ππ)2
2ππ
π‘
=4π2ππ‘
2ππ
π½ = 2ππ3π‘
Unsymmetrical Bending:-
If the cross section of the beam is not symmetrical about any axis or applied load is not acting
through plane of symmetry than bending will be unsymmetrical
Consider a beam obituary cross section as shown in fig,
Internal Force System
Resolution of bending moments
Sign depending on the size of π.In both cases, for the sense of M shown
ππ₯ = ππ πππ
ππ¦ = ππππ π
Which give,
For π <π
2, ππ₯ and ππ¦ positive (fig (a)) and for π >
π
2, ππ₯ positive and ππ¦ negative (fig
(b)).
If the neutral axis made angle πΌ with x-axis
π¦β² = π₯π πππΌ + π¦πππ πΌ
Strain
π =π¦β²
π
π =π₯π πππΌ+π¦πππ πΌ
π
ππ§ = πΈπ
ππ₯ =πΈ
π (π₯π πππΌ + π¦πππ πΌ)
Strain relations
β« ππ§ππ΄ = 0
(zero about Neutral axis)
Moment resultant
ππ₯ = β« ππ§ π¦ππ΄
ππ¦ = β« ππ§π₯ππ΄
ππ₯ =πΈ
π [π πππΌ β« π₯π¦ππ΄ + πππ πΌ β« π¦2ππ΄]
=πΈ
π [π πππΌπΌπ₯π¦ + πππ πΌπΌπ₯π₯]
Similarly
ππ¦ =πΈ
π [πππ πΌπΌπ₯π¦ + π πππΌπΌπ¦π¦]
This is matrix form
[ππ₯
ππ¦] =
πΈ
π [πΌπ₯π₯ πΌπ₯π¦
πΌπ₯π¦ πΌπ¦π¦] {
πππ πΌπ πππΌ
}
{πππ πΌπ πππΌ
} =πΈ
π [ππ₯
ππ¦] [
πΌπ₯π₯ πΌπ₯π¦
πΌπ₯π¦ πΌπ¦π¦]
β1
{πππ πΌπ πππΌ
} =π
πΈ(πΌπ₯π₯πΌπ¦π¦βπΌπ₯π¦2)
[πΌπ¦π¦ βπΌπ₯π¦
βπΌπ₯π¦ πΌπ¦π¦] {
ππ₯
ππ¦}
πππ πΌ =π
πΈ(πΌπ₯π₯πΌπ¦π¦βπΌπ₯π¦2)
(πΌπ¦π¦ππ₯ β πΌπ₯π¦ππ¦)
π πππΌ =π
πΈ(πΌπ₯π₯πΌπ¦π¦βπΌπ₯π¦2)
(βπΌπ₯π¦ππ₯ + πΌπ₯π₯π)
ππ§ =πΈ
π (π₯π πππΌ + π¦πππ πΌ)
ππ§ =(πΌπ₯π₯ππ¦βπΌπ₯π₯ππ₯)
(πΌπ₯π₯πΌπ¦π¦βπΌπ₯π¦2)
π₯ +(πΌπ¦π¦ππ₯βπΌπ₯π¦ππ¦)
(πΌπ₯π₯πΌπ¦π¦βπΌπ₯π¦2)
π¦
ππ§ =ππ¦
πΌπ¦π¦π₯ +
ππ₯
πΌπ₯π₯π¦
At ππ¦ = 0
ππ§ =ππ₯π¦
πΌπ₯π₯
Problems:-
1). An idealized thin walled cross-section of the beam and perspective areas of boom are as
shown in a bending moment ππ¦ is acting on the cross-section the ratio of magnitude of normal
stress in the top boom that of bottom boom.
(a). 5
11
(b). 2
5
(c). 1
(d). 5
2
Sol:-
Since it is symmetric
ππ₯ =ππ₯
πΌπ₯π₯π¦ +
ππ¦
πΌπ¦π¦π₯
Since, ππ₯ = 0( No bending stress at x-axis )
ππ§ =ππ¦
πΌπ¦π¦π₯
ππ§ π‘ππ =ππ¦
πΌπ¦π¦π₯π‘ππ; ππ§ πππ‘π‘ππ =
ππ¦
πΌπ¦π¦π₯πππ‘π‘ππ
ππ§ π‘ππ
ππ§ πππ‘π‘ππ=
ππ¦
πΌπ¦π¦π₯π‘ππ
ππ¦
πΌπ¦π¦π₯πππ‘π‘ππ
ππ§ π‘ππ
ππ§ πππ‘π‘ππ=
π₯π‘ππ
π₯πππ‘π‘ππ
π₯π‘ππ = 2 β οΏ½Μ οΏ½
π₯πππ‘π‘ππ = 2 + οΏ½Μ οΏ½
οΏ½Μ οΏ½ =β π΄οΏ½Μ οΏ½
β π΄=
2 π 2 π 0.2+0.1 π 2+0β0.2 π 2
3 π 0.2+2 π0.1=
0.8+0.2β0.4
0.6+0.2
οΏ½Μ οΏ½ = 0.75 =3
4
π₯π‘ππ = 2 β3
4=
16
4
ππ§ π‘ππ
ππ§ πππ‘π‘ππ=
π₯π‘ππ
π₯πππ‘π‘ππ=
5
411
4
=5/11