beam deflections using double integration double...example problem a w x y #$ modulus of elasticity...
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Beam Deflections Using Double IntegrationStevenVukazich
SanJoseStateUniversity
π"π¦ππ₯"
=ππΈπΌ
Recall the Moment-Curvature Relationshipfor Small Deformations
In order to solvethis differential equation for y and π we need:β’ Moment equation (from statics);β’ Two boundary (or continuity)
conditions on y or π;β’ Information on E and I.
PM
w
x
y π π₯ =ππ¦ππ₯
π¦ π₯
Example Problem
A
w
x
y
π¦ π₯Modulus of Elasticity = EMoment of Inertia = I
B
Find the equation of the elastic curve for the simply supported beam subjected to the uniformly distributed load using the double integration method. Find the maximum deflection. EI is constant.
L
Free Body Diagram of the Beam
Need to find the moment function M(x)
Ax
Ay By
A
w
x
y
B
L
0π1
οΏ½
οΏ½
= 0+
Free Body Diagram of the Beam
Ax
Ay By
A
w
x
y
B
L
12π€πΏ
23πΏ
π΅; =13π€πΏ
Free Body Diagram of the Beam
Ax
Ay By
A
w
x
y
B
L
12π€πΏ
23πΏ
π΄; =16π€πΏ
0πΉ;
οΏ½
οΏ½
= 0+
0πΉ?
οΏ½
οΏ½
= 0+
Free Body Diagram of the BeamShowing Known Reactions
A
w
x
y
B
L
13π€πΏ
16π€πΏ
a
a
Cut the beam at a-a to find the moment function M(x)
x
M
V16π€πΏ
π€ =π€πΏπ₯
Free Body Diagram of Segment to the Left of aβa
x
M
V16π€πΏ
π =12π₯
π€πΏπ₯ =
π€2πΏπ₯"
23π₯
13π₯
a
a
a
a 0πA
οΏ½
οΏ½
= 0+
π =π€πΏ6π₯ β
π€6πΏπ₯C
Need Two Boundary Conditions on y or π
A
w
x
y
π¦ π₯Modulus of Elasticity = EMoment of Inertia = I
B
L
π¦ 0 = 0
π¦ πΏ = 0
Solve the Differential Equation
A
w
x
y
π¦ π₯Modulus of Elasticity = EMoment of Inertia = I
B
L
π¦ 0 = 0
π¦ πΏ = 0π"π¦ππ₯"
=ππΈπΌ
π =π€πΏ6π₯ β
π€6πΏπ₯C
For constant EI
πΈπΌπ"π¦ππ₯"
= π
πΈπΌπ"π¦ππ₯"
=π€πΏ6π₯ β
π€6πΏπ₯C
πΈπΌπ"π¦ππ₯"
=π€πΏ6π₯ β
π€6πΏπ₯C
First Integration
DπΈπΌπ"π¦ππ₯"
οΏ½
οΏ½
ππ₯ = Dπ€πΏ6π₯ β
π€6πΏπ₯C ππ₯
οΏ½
οΏ½
πΈπΌππ¦ππ₯
= βπ€24πΏ
π₯F +π€πΏ12
π₯" + πΆH
πΈπΌπ = βπ€24πΏ
π₯F +π€πΏ12
π₯" + πΆH
π¦ 0 = 0
π¦ πΏ = 0
Second Integration
DπΈπΌππ¦ππ₯
οΏ½
οΏ½
ππ₯ = D βπ€24πΏ
π₯F +π€πΏ12
π₯" + πΆH ππ₯οΏ½
οΏ½
πΈπΌππ¦ππ₯
= βπ€24πΏ
π₯F +π€πΏ12
π₯" + πΆH
πΈπΌπ¦ = βπ€
120πΏπ₯I +
π€πΏ36
π₯C + πΆHπ₯ + πΆ"
Use the two boundary conditions to find C1 and C2
Find Constants C1 and C2
π¦ 0 = 0
π¦ πΏ = 0
πΈπΌπ¦ = βπ€
120πΏπ₯I +
π€πΏ36
π₯C + πΆHπ₯ + πΆ"
πΈπΌ 0 = βπ€
120πΏ0 I +
π€πΏ36
0 C + πΆH 0 + πΆ"
πΈπΌ 0 = βπ€
120πΏ0 I +
π€πΏ36
0 C + πΆH 0 + πΆ"
πΆ" = 0
πΈπΌ 0 = βπ€
120πΏπΏ I +
π€πΏ36
πΏ C + πΆH πΏ + πΆ"
πΆH =π€πΏC
120βπ€πΏC
36= β
7π€πΏC
360
Functions for y and π
πΈπΌπ¦ = βπ€
120πΏπ₯I +
π€πΏ36
π₯C β7π€πΏC
360π₯
πΈπΌπ = βπ€24πΏ
π₯F +π€πΏ12
π₯" β7π€πΏC
360
π¦ π₯ =1πΈπΌ
βπ€
120πΏπ₯I +
π€πΏ36
π₯C β7π€πΏC
360π₯
π π₯ =1πΈπΌ
βπ€24πΏ
π₯F +π€πΏ12
π₯" β7π€πΏC
360
Functions for y and π
A
w
x
y
π¦ π₯Modulus of Elasticity = EMoment of Inertia = I
B
L
π¦ π₯ =1πΈπΌ
βπ€
120πΏπ₯I +
π€πΏ36
π₯C β7π€πΏC
360π₯
π π₯ =1πΈπΌ
βπ€24πΏ
π₯F +π€πΏ12
π₯" β7π€πΏC
360
π π₯ =ππ¦ππ₯
Questions
A
w
x
y
π¦ π₯
Modulus of Elasticity = EMoment of Inertia = I
B
L
π¦ π₯ =1πΈπΌ
βπ€
120πΏπ₯I +
π€πΏ36
π₯C β7π€πΏC
360π₯
π π₯ =1πΈπΌ
βπ€24πΏ
π₯F +π€πΏ12
π₯" β7π€πΏC
360
π π₯ =ππ¦ππ₯
How can we find the maximum displacement?Where does the maximum displacement occur?How can we find the rotation at the supports?
Answers
A
w
x
y
π¦max
Modulus of Elasticity = EMoment of Inertia = I
B
Lπ¦ π₯ =
1πΈπΌ
βπ€
120πΏπ₯I +
π€πΏ36
π₯C β7π€πΏC
360π₯
π π₯ =1πΈπΌ
βπ€24πΏ
π₯F +π€πΏ12
π₯" β7π€πΏC
360
π = 0
ymax occurs where π = 0
πA is π(0) πB is π(L)
?
π πΏ
π 0
Point Where Maximum Deflection Occurs
π =βπ Β± π" β 4πποΏ½
2π
π π₯ =1πΈπΌ
βπ€24πΏ
π₯F +π€πΏ12
π₯" β7π€πΏC
360= π
Set π = 0
β124π" +
πΏ"
12π β
7πΏF
360= π
Let π = π₯" Multiply both sides by TUVW
π =β πΏ"12 Β±
πΏF144 β 4 β 1
24 β 7πΏF360
οΏ½
2 β 124
= πΏ" Β± 12πΏF
144β
16
7πΏF
360οΏ½
= πΏ" Β± 12πΏ"1270
οΏ½
π = πΏ" Β± 4πΏ"130
οΏ½= 1 + 4
130οΏ½ , π β π
ππποΏ½ πΏ"
π₯ = 1 β 4130οΏ½
οΏ½πΏ = 0.51893πΏ
Maximum Deflection
A
w
x
y
π¦max
Modulus of Elasticity = EMoment of Inertia = I
B
L
π = 0
π πΏ
π 0
1 β 4130οΏ½
οΏ½πΏ = 0.51893πΏ
π¦max =1πΈπΌ
βπ€
120πΏ0.5193πΏ I +
π€πΏ36
0.5193πΏ C β7π€πΏC
3600.5193πΏ = β0.006522
π€πΏF
πΈπΌ
Slope at Supports
A
w
x
y
Modulus of Elasticity = EMoment of Inertia = I
B
L
π1 = π 0 =1πΈπΌ
β7π€πΏC
360= β
7π€πΏC
360πΈπΌ= β0.01944
π€πΏC
πΈπΌ
π πΏ
π 0
π` = π πΏ =1πΈπΌ
βπ€24πΏ
πΏF +π€πΏ12
πΏ" β7π€πΏC
360=π€πΏC
45πΈπΌ= 0.02222
π€πΏC
πΈπΌ
Compare to Tabulated Solution in AISC Manual
π =12π€πΏ