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)