direct integration of differential equation of an elastic
TRANSCRIPT
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)