Access over 20 million homework & study documents
search

Castigliano

Content type

User Generated

Subject

Engineering

Type

Homework

Rating

Showing Page:
1/6
Q1.
Consider a section X-X at a distance x in from A as shown in figure A
The moment equation for section CD
 
is given by
  
Applying virtual load at A where deflection is to be determined, we will have a section as shown
in Figure B
 )
We will have the following equation

Therefore, the deflection at A is given by;


Substituting, we have 











(downward)

Sign up to view the full document!

lock_open Sign Up
Showing Page:
2/6
Q2.
We start by solving support reactions as shown in figure C
 



   
For deflection,
Member
Origin
Limits
BC
B
0.225
= 1.01x
0
CA
C
0 1.35

-x
To obtain the deflection at C, we apply a virtual load of 1 kip at C as shown in Figure D



Sign up to view the full document!

lock_open Sign Up
Showing Page:
3/6

Sign up to view the full document!

lock_open Sign Up
End of Preview - Want to read all 6 pages?
Access Now

Unformatted Attachment Preview

Q1. Consider a section X-X at a distance x in from A as shown in figure A The moment equation for section CD (2 𝑖𝑛 < 𝑥 < 6 𝑖𝑛) is given by𝑀𝑥 = −4(𝑥 − 2) Applying virtual load at A where deflection is to be determined, we will have a section as shown in Figure B (2 𝑖𝑛 < 𝑥 < 6𝑖𝑛) We will have the following equation 𝑚𝑥 = −𝑥 6 𝑚𝑥 𝑀𝑥 Therefore, the deflection at A is given by; 𝛿 = ∫2 Substituting, we have 𝐸𝐼 𝑑𝑥 6 (−𝑥)(−4(𝑥−2)) 𝛿 = ∫2 𝐸𝐼 6 4𝑥 2 − 8𝑥 = ∫ 𝑑𝑥 𝐸𝐼 2 = 149.333 ↓ 𝐸𝐼 (downward) 𝑑𝑥 Q2. We start by solving support reactions as shown in figure C 𝑉𝐴 + 𝑉𝐵 = 4 ∗ 1.35 = 5.4 𝑘𝑖𝑝 ∑ 𝑀𝐴 = 0 1.35 𝑉𝐵 ∗ 3.6 = 4(1.35) ( ) 2 𝑉𝐵 = 1.01 𝑘𝑖𝑝 𝑉𝐴 = 5.4 − 1.01 = 4.39 𝑘𝑖𝑝 For deflection, Member BC CA Origin B C 𝑀𝑥 ↑ 𝑉𝐵 𝑥 = 1.01x Limits 0.225 0 – 1.35 (1.01 ∗ 2.25) − 4 ∗ 𝑀1 𝑥2 2 0 -x To obtain the deflection at C, we apply a virtual load of 1 kip at C as shown in Figure D 𝐿 ∆= ∫ 0 𝑀𝑥 𝑚 𝑑𝑥 𝐸𝐼 2.25 1.35 0 0 1 ∆𝐶 = ∫ (1.01)𝑥(0)𝑑𝑥 + ∫ [(2.2725) − 2𝑥 2 ]−𝑥 𝑑𝑥 𝐸𝐼 1.35 1 ∆𝐶 = ∫ (−2.2725𝑥 + 2𝑥 3 )𝑑𝑥 𝐸𝐼 0 ∆𝐶 = − 0.41 𝑖𝑛 𝐸𝐼 Q3. Solving the end reactions, 𝑅𝐴 = 𝑅𝐵 = 2+(2∗8) 2 = 9 𝑘𝑖𝑝 𝑑2𝑦 ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

Anonymous
I use Studypool every time I need help studying, and it never disappoints.

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4