Mechanics of Aircraft Structures (3rd Edition, 2026) – Solutions Manual – by Sun

Solutions Manual For Mechanics Of
Aircraft Structures (3rd Edition, 2026)
–By C.T. Sun And Ashfaq Adnan
Newest Version

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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
L = 2(A + B) (See Fig. 1.16) Is Fixed, Find The Optimum B/A Ratio To Achieve The
Most Efficient Section If M = T And  Allowable = 2 Allowable . Note That For Closed
Thin-Walled Sections Such As The One In Fig.1.16, The Shear Stress Due To Torsion
Is
T
=
2abt




Figure 1.16 Closed Thin-Walled Section


Solution:
My
(1) The Bending Stress Of Beams  = , Where Y Is The Distance From The Neutral
Is I

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

1
With Height H And Width I= ( Wh 3 + Ad 2 In Which A Is The
W. 12 )

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
1 1 B 2 Tb
Result I = 2  Tb + 2 [  At + (At)  ) ] 2 (3a + B) , Assuming That T Is
Is 12 3 12 3 ( 
2 6


1.1.1




Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

,Name: Mohamed Naleer Abdul Gaffor Email: IP: 184.162.144.24


Mechanics of Aircraft structures
C.T. Sun
Very Small.
 Allowable First And Find
(2) The Shear Stress Due To Torsion For A Closed Thin-
Walled Section Shown Above Is
T
= .
2abt

(3) Two Approaches Are Employed To Find The Solution.
(i) Assume That The Bending Stress Reaches The
Allowable
The Corresponding Bending Maximum Bending Moment. Then Apply The Stated
Loading Condition T = M To Check Whether The Corresponding  Max Has
Of
Exceeded The Allowable Shear  Allowable . If This Condition Is Violated, Then
Stress The Optimized B/A Ratio Is
Not Valid.
B
M
My 2 3M
(A)  | Y = = = =
I Tb Tb(3a +
B
2 2 (3a + B)
B)
6
When Given L = 2(A + B) As A Constant, A Can Be Expressed In Terms Of
B
L
And L A= − B . Then We Can Minimize
As 2

Tb(3a + B) Tb(3L − In Order To Maximize  , I.E.,
=
S4b) =
3 6
S T 3L L L
= 0 (3L − 8b) = 0  B = , So A = −B=
B 6 8 2 8
B
Where The Optimum =3
Ratio Is A

3M 3M 32M
Thus,  Max = = =
Tb(3a + T  (3L / 8)  (3  L / 8 + 3L / 3tl2
B) 8)
(B)  Max Wit T = M And B/A = 3 And Check  Max Is Within
Check h Whether
The Allowable Shear  Allowable .
Stress
T M 32M
 Max = = = =  Max =  Allowable
2abt 2  (L / 8)  (3L / 8)  3tl2
T
 Allowable
  Allowable =
2
The Result Above Means That Under This Assumption, Shear Stress Would
Reach The Allowable S tress  Allowable
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

,Name: Mohamed Naleer Abdul Gaffor Email: IP: 184.162.144.24


Mechanics of Aircraft structures
C.T. Sun



Before  Reaches 
.
Allowable
Conseq
uently,
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
 Max =  Is Reached First And Then Find The Corresponding Bending Stress
Allowable

According To The Loading Condition M = T .
T
(A)  =
2abt
Again We S = 2abt = (L − 2b)Bt In Order To Maximize , I.E.,
Minimize


1.1.2





Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN