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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