Mechanics of Solids & Aircraft Structural Analysis Assignment 2

This document presents a detailed solution to Assignment 2 for the ME4226 Mechanics of Solids 2 and ME4063 Aircraft Structural Analysis courses. The solution addresses two main questions. Question 1 focuses on strain analysis, involving the construction of Mohr's strain circle using strain gauge data from a 7075-T6 Aluminium Alloy component. The solution determines principal strains and stresses, the angle between strain components, and assesses yielding using the von Mises stress criterion. Question 2 involves the analysis of beam deflection and stress. It requires the calculation of the second moment of area for different beam profiles, determining deflection and maximum stress at a point, and identifying the location of maximum deflection using Macauley's method. The solution includes all necessary calculations and formulas to solve the problems.

Question 1
Part a
ε x=εa=1142 με
ε y=ε c=−140 με
ε b=458 με
r xy=2 ε b−ε a−ε c
¿ 2 ( 458 ) −1142+140
¿−86 με
Centre of Mohr’s circle
C= [ ε x+ ε y
2 ,0 ]= [ 1142−140
2 , 0 ]= [501 , 0 ]
[501, 0]

Radius of Mohr’s circle
R=CA=CB=CM
R= √ ( 1142−501 ) 2+ 432=642.441
Principal strain
ε 1=OC + R=501+642.441=1143.441 με
ε 2=− [ R−OC ]=− [ 642.441−501 ]=−141.441 με
Part b
The angle that ε y or ε x makes with ε 1
r xy
2
R =sin 2θp
θp =1
2 [ sin−1 43
642.441 ] =1.920
Part c
σ 1= E
1−v2 [ ε2+ v ε1 ]= 71.7∗103
1−0.332 [ 1143.441+ ( 0.33∗−141.441 ) ]∗10−6
σ 1=88.248 MPa

σ 2= E
1−v2 [ ε1+ v ε2 ]= 71.7∗103
1−0.332 [−141.441+ ( 0.33∗1143.441 ) ]∗10−6
σ 2=18.981 MPa
Part d
σ von= √ σ 1
2+σ2
2−σ 1 σ 2= √88.2482+ 18.9812− ( 88.248∗18.981 )
σ von=80.455 MPa
Part e
σ von <σ y ( 503 MPa )
Yielding will not take place.
Question 2
Part a
a) Second moment of area ¿ b h3
12
I = [ w h3
12 ]−2 [ ( w−2 t ) ( h
2 −3
2 t )3
12 ]−2 [ ( h−2 t ) ( w
2 − t
2 )3
12 ]
I = 0.06∗0.193
12 −2 [ 0.036∗0.0773
12 ]−2 [ 0.146∗0.04463
12 ]
I =3.9113∗10−5 m4
b) Second moment of area
I = [ w h3
12 ] −2 [ b ( h−t ) 3
12 ] −[ ( w−2 b ) ( h−t ) 3
12 ]
I = [ 0.1∗0.173
12 ] −2 [ 0.025∗0.1583
12 ] −[ 0.05∗0.1583
12 ]
I =8.0724∗10−6 m4
Part b

Calculating the Reaction Forces (R¿¿ 1∧R2) ¿
∑ Fv=0
⇒ R1−P1 −W (0.6 m)+ R2−P2=0
⇒ R1−5000−6000(0.6)+ R2 −2000=0
⇒
R1 +R2=10600 N ……..(1)
Moments at (R2 )
∑ M R2
=0
⇒ R1 (2.4)−5000(1.8)−[6000 ( 0.6
2 +0.6)( 0.6)]−2000(1.6)=0
⇒
2.4 R1 =9000+3240 ⎼ 3200
⇒ R1=3766.667 N Substitute R1 in Eq (1).
R2=10600 ⎼ 3766.667
⇒ R2=6833.34 N .
 Deflection of the beam at x=1.84 m, can be obtained by using Macauley’s Method:
⎼ EI d2 v
d x2 =M
 Write the general formula of moment equation for the last right-hand side portion of the
beam.
M x=3766.667 ( x )−5000 ( x−0.6 )− 6000
2 ( x−1.2 )2 + 6000
2 ( x−1.8 )2+¿6833.33 ( x−2.4 )
By successive integration of this equation (Using Macauley’s integration).
−EI dv
dx =3766.667 ( x )2
2 −5000 ( x−0.6 )2
2 −6000 ( x−1.2 )3
6 + 6000 ( x−1.8 )3
6 + 6833.33 ( x−2.4 )2
2 +C1
 Second Integration, (The general deflection equation).

−EIV =3766.667 ( x ) 3
6 −5000 ( x−0.6 ) 3
6 − 6000 ( x−1.2 ) 4
24 + 6000 ( x−1.8 ) 4
24 + 6833.33 ( x−2.4 ) 3
6 +C1 x +C2
 Apply the Boundary condition, (B.C is usually being applied at the positions where the deflection isn’t
functioning).
i. First B.C V =0@ x=0 m Substitute in General equation.
−EI (0)=0−0+0+ 0+C1 (0)+C2 ⇒ C2=0
ii. Second assumption V =0@ x=2.4 m Substitute in General equation.
0=3766.667 ( 2.4 ) 3
6 −5000 ( 2.4−0.6 )3
6 − 6000 ( 2.4−1.2 ) 4
24 + 6000 ( 2.4−1.8 ) 4
24 +C1 (2.4)+0
⇒ 0=3332.400+2.4 C1 ⇒ ⎼ C1= 3332.400768
4 ⇒C1 =⎼ 1388.500
 Deflection in the beam of x=2.15 m, (Substitute both C1& x into the general deflection equation, (Note:
The remaining terms can be neglected as they are negative).
−EIV =3766.667 ( 1.84 ) 3
6 − 5000 ( 1.84−0.6 ) 3
6 − 6000 ( 1.84−1.2 ) 4
24 + 6000 ( 1.84−1.8 ) 4
24 −(1.84)
 ⇒ ⎼ EIV =2279.9488 ⎼ 2554.84=274.8912 ⇒ V =−274.8912
EI C1& x into the general deflection
equation, (Note: The remaining terms can be neglected as they are negative).
−EIV =3766.667 ( 1.84 )3
6 − 5000 ( 1.84−0.6 )3
6 − 6000 (1.84−1.2 )4
24 + 6000 ( 1.84−1.8 )4
24 −1388.500(1.84)
V = 274.888
EI
Aluminium: E=71.7 GN
m2 , Ia=3.9113∗10−5 m4 , Ib=8.0724∗10−6 m4
For beam a
V = 274.888∗10−3
71.7∗103∗3.9113∗10−5 =0.098 m
For beam b
V = 274.888∗103
71.7∗103∗8.0724∗10−6 =0.475 m
Part c

σ = EV
L
For beam a
σ = ( 71.7∗103∗0.098 )
1.24 =5666.613 N /m m2
For beam b
σ = ( 71.7∗103∗0.475 )
1.24 =27465.725 N /m m2
Part d
At maximum deflection, dV
dx =0
−EI ( 0 ) = 3766.667 ( x ) 2
2 − 5000 ( x−0.6 ) 2
2 −6000 ( x−1.2 ) 3
6 + 6000 ( x−1.8 )3
6 + 6833.33 ( x−2.4 ) 2
2 −1388.50
Solving for x,
x=2.353 m
Maximum deflection occur at 2.353 m from support 1.