Direct integration of differential equation of an elastic ...
Transcript of Direct integration of differential equation of an elastic ...
Direct integration of differential equation
of an elastic curve to determine the
slope and the deflection of a beam
w = deflection
w´ = φ = slope
EIw´´ = -M basic differential equation
EIwIII = -V
EIwIV = q
Schwedlers relations:
Deformations at bending:
1. Slope [rad]
2. Deflection [m] - w
Simple beam
Cantilever beam
w = 0 w = 0
w = 0
w´= φ = 0
wmax in the place of w´(x) = 0
(we will get points of inflexion on the
elastic curve, real root is the place of
wmax)
– this is not the same place as Mmax
wmax
w´(x) = 0
x
wmax
φAφB
Deformation (boundary)
conditions in supports
M
V
M
V
φmax
Wmax is always at the end of the
beam (points of inflexion are not in
this´case place of the wmax )
Example 1Determine the slope and deflection of the free end b:
Ma= F.lF
ba
x
l
Ra=F
w =0w´=0
Boundary conditions: 1. w´(x=0)=0: C1=02. w(x=0)=0: C2=0
Equation of slope: 𝑤 =1
𝐸𝐼∙ 𝐹 ∙ 𝑙 ∙ 𝑥 − 𝐹 ∙
𝑥2
2
Equation of deflection: 𝑤 =1
𝐸𝐼∙ 𝐹 ∙ 𝑙 ∙
𝑥2
2− 𝐹 ∙
𝑥3
6
Slope at b: 𝜑𝑏 = w´(x=l) =1
𝐸𝐼∙ 𝐹 ∙ 𝑙 ∙ 𝑙 − 𝐹 ∙
𝑙2
2=
𝐹∙𝑙2
2𝐸𝐼
Deflection at b: wb = w(x=l) = 1
𝐸𝐼∙ 𝐹 ∙ 𝑙 ∙
𝑙2
2− 𝐹 ∙
𝑙3
6=
𝐹∙𝑙3
3𝐸𝐼
Equation of the elastic curve and double integration :
𝑀(𝑥) = −𝑀𝑎 + 𝑅𝑎 ∙ 𝑥
𝑀(𝑥) = −𝐹 ∙ 𝑙 + 𝐹 ∙ 𝑥
𝐸𝐼 ∙ 𝑤´´ = −𝑀(𝑥)
𝐸𝐼 ∙ 𝑤´´ = +𝐹 ∙ 𝑙 − 𝐹 ∙ 𝑥
𝐸𝐼 ∙ 𝑤´ = +𝐹 ∙ 𝑙 ∙ 𝑥 − 𝐹 ∙𝑥2
2+𝐶1
𝐸𝐼 ∙ 𝑤 = +𝐹 ∙ 𝑙 ∙𝑥2
2− 𝐹 ∙
𝑥3
6+𝐶1 ∙ 𝑥 + 𝐶2
Example 2
F
l
w =0w´=0
a b
M(x) = -F.x
EI.w´´= -M(x)
EIw´´= +F.x
EIw´=+F.x2/2 + C1
EIw = +Fx3/6 + C1x + C2
Boundary conditions: 1. w´(x=l)=0: C1=-F.l2/2 2. w(x=l)=0: +Fl3/6 + C1l + C2 = 0
+Fl3/6 -F.l3/2 + C2 = 0C2=-F.x3/3
Equation of slope: w´=1/EI (F.x2/2 - F.l2/2 )Equation of deflection: w =1/EI (F. x3/6 +F.l2.x/2+F.l3/3)Slope at a: a= w´(x=0) = C1 = Fl2/(2EI)Deflection at a: wa= w(x=0) = C2 = Fl3/(3EI)
Determine the slope and deflection of the free end a:
x
l
q
M(x) = Ra.x - q.x2/2
M(x) = q.l/2 .x - q.x2/2
EI.w´´= -M(x)
EI.w´´ = -q.l/2 .x +q.x2/2
EI.w´= -q.l/2 .x2/2 + q.x3/6 + C1
EI.w = -q.l/2 .x3/6 + q.x4/24 + C1. x + C2
Boundary conditions: 1.w(x=0)=0: C2=0
2.w(x=l)=0: C1= ql3/24
Slope at a: a= w´(x=0) = ql3/(24EI)
Slope at b: wc = w(x=l) = -ql3/(24EI)
Deflection at l/2: wmax = w (x=l/2) = 5/384 .ql4/EI
Ra= q.l/2
x
w = 0w = 0
a b
lim
4
max384
5w
EI
lqw k
Example 3
Example 4
M(x) = -q.x2/2
EI.w´´= -M(x)
EIw´´= q.x2/2
EIw´= qx3/6 + C1
EIw = qx4/24 + C1x + C2
Boundary conditions:
1.w´(x=l)=0: C1=- ql3/6 2.w(x=l)=0: ql4/24 + C1.l + C2 = 0
ql4/24 + - ql4/6 + C2 = 0
C2=ql4/8
Slope at a: a=w´ (x=0)= C1=- ql3/(6EI)
Deflection at a: wa=w (x=0)=C2=(ql4/8EI)
q
ba
w = 0w´=0
l
Determine the slope and deflection of the free end a:
x
l
M(x) = -Ra.x
M(x) = -M/l .x
EI.w´´= -M(x)
EI.w´´ = M/l .x
EI.w´= Mx2/2l + C1
EI.w = Mx3/6l + C1. x + C2
Boundary conditions: 1.w(x=0)=0: C2=0
2.w(x=l)=0: C1= -Ml/6
Equation of slope: w´ = 1/EI (Mx2/2l -Ml/6)
Equation of deflection: w = 1/EI (Mx3/6l -Ml/6x)
Slope at a: a= w´(x=0) = -Ml/(6EI)
Slope at b: b = w(x=l) = Ml/(3EI)
Deflection at l/2: w (x=l/2) = -Ml2/(24EI)
Ra= M/l
x
w = 0
w = 0
a b
Example 5
M
The position of maximal deflection: at x: when w´ (x) = 0
w´= 1/EI (Mx2/2l - Ml/6)
Mx2/2l - Ml/6= 0
x =l2
3
l
M(x) = Ra.x - M
M(x) = M/l .x - M
EI.w´´= -M(x)
EI.w´´ = -M/l .x + M
Boundary conditions: 1.w(x=0)=0:
2.w(x=l)=0:
Slope at a: a= w´(x=0)
Slope at b: b = w(x=l)
Deflection at l/2: w (x=l/2)
Ra= M/l
x
w = 0 w = 0
a b
Example 6
M
Example 7Determine the slope and deflection of the free end b:
M(x) = -Ma + Ra.x – qx2/2
M(x) = -q.l2/2 +q.l.x– qx2/2
EI.w´´= -M(x)
EI.w´´ = +q.l2/2 -q.l.x+ qx2/2
Ma= q.l2/2 q
ba
x
l
Ra=q.lw =0w´=0
Boundary conditions: 1. w´(x=0)=0: C1=02. w(x=0)=0: C2=0
Slope at b: b=w´ (x=0)= C1=- ql3/(6EI)Deflection at b: wb=w (x=0)=C2=(ql4/8EI)