Chapter 9 Deflection of Beams and Shafts

Method of Superposition

Method of Superposition

P w

M

A

1. The loading can be separated into three parts:

2. For each part, the displacement and slope at A can be found using the table in Appendix C (pp. 853-854)

w

A

P

A

M

A

3. The total displacement and slope at A are the algebraic sums of these parts.

L/2 L/2

Method of Superposition

12-89 The W200-71 cantilevered beam is made of A-36 steel and is subjected to the loadingshown. Determine the displacement at its end A. (p. 660)

6 kN

2.4 m 2.4 m

3 kN m

A

B

P

A

M

A

1. The loading can be separated two parts:

2. For each part, the displacement can be found using the table in Appendix C

First Part, P:

Second Part, M:

(L=2.4 x 2 =4.8 m)

Method of Superposition

12-89 The W200-71 cantilevered beam is made of A-36 steel and is subjected to the loadingshown. Determine the displacement at its end A. (p. 660)

M

A

2. For each part, the displacement can be found using the table in Appendix C

First Part:

Second Part:

(L=2.4 x 2 =4.8 m)

BL/2

B

AѲB

L/2

3. The total displacement at A:

=16.13 mm

Method of Superposition

12-100 Determine the vertical deflection and slope at the end A of the bracket. EI isconstant. (p. 662)

75

mm

100 mm400 N

4 kN/m

B

ѲB_1

A

400 N

AB ѲB_1

1. The loading can be separated two parts:

First loading Part, P:

(LBC=75 mm)

(LAB=100 mm)

C

C

400 NB

C

ѲB_1

BC :

AB:

Method of Superposition

12-100 Determine the vertical deflection and slope at the end A of the bracket. EI isconstant. (p. 662)

B

ѲB_2

A

ѲB_2

First loading Part, P:

C

4 kN/m

Second loading Part, w:

B

C

PB=0.4 kN

MB=w(LAB)2/2=20 N m

BC :

AB:

The total displacement and slope at A:

Method of Superposition

12-100 Determine the vertical deflection and slope at the end A of the bracket. EI isconstant. (p. 662)

B

ѲB_2

A

ѲB_2

C

4 kN/m

B

C

PB=0.4 kN

MB=w(LAB)2/2=20 N m

The total displacement and slope at A:

First loading Part, P:

Second loading Part, w:

Chapter 9 Deflection of Beams and Shafts

Statically Indeterminate Beams

Statically Indeterminate Beams

Superposition Method

P

L/2 L/2A

FA_x

FA_y

MA

Equilibrium conditions:

B

FB_y

Compatibility Conditions

P

FB_y

Deflection at x=L/2

xv

C

vc

Statically Indeterminate Beams

Double Integration Method P

L/2 L/2A

FA_x

FA_y

MA

B

Step 1 Moment functions

P

L/2 L/2A

FA_x

FA_y

MA

B

FB_y

v=0

FB_yL/2

MA

M(x)

x

Step 2 Double integration for the elastic equation

2.1 Integrate once

2.2 Integrate twice

2.3 Boundary Conditions

Compatibility Conditions

(At point A: Ѳ=0; v=0)

At x=L; Ѳ=0

At x=L; v=0

At x=0; v=0

C

Statically Indeterminate Beams

Double Integration Method P

L/2 L/2A

FA_x

FA_y

MAB

P

L/2 L/2A

FA_x

FA_y

MA

B

FB_y

v=0

FB_yL/2

MA

M(x)

x

Step 2 Double integration for the elastic equation

2.4 The equation for elastic curve

Deflection at x=L/2

C

Statically Indeterminate Beams

Superposition Method

P

L/2 L/2A

FA_x

FA_y

MA

Equilibrium conditions:

B

Compatibility Conditions

P

FB_y

C

F’B_y

C

Bε

(Example 12.23 p.681)

Statically Indeterminate Beams

Superposition Method

12-125 Determine the reactions at the support C. EI is constant. (p. 686)

PD

B

A C

L/2L/2

A C

F’B_y

PD

B

FB_y

1. Equilibrium conditions:

2. Superposition principle:

3. Compatibility conditions:

Symmetry condition:

Segment DB

Segment AC

Statically Indeterminate Beams

Superposition Method

12-124 Determine the reactions at the supports A, B and C, then draw the shear and momentdiagram. EI is constant. (p. 685)

60 kN

AB

C

50 kN/m

2 m 2 m 4 m

A C

w=50 kN/m

P=60 kN

A C

A C

FB_y

1. Equilibrium conditions:

2. Superposition principle:

Statically Indeterminate Beams

Superposition Method

12-124 Determine the reactions at the supports A, B and C, then draw the shear and momentdiagram. EI is constant. (p. 685)

60 kN

AB

C

50 kN/m

2 m 2 m 4 m

A C

w=50 kN/m

P=60 kN

A C

A C

FB

2. Superposition principle:

3. Compatibility conditions:

Statically Indeterminate Beams

Superposition Method

12-124 Determine the reactions at the supports A, B and C, then draw the shear and momentdiagram. EI is constant. (p. 685) 60 kN

AB

C

50 kN/m

2 m 2 m 4 m

3. Compatibility conditions:

Equilibrium conditions:

x

V

11.88 kN

-48.13 kN

118.12 kN

-81.88 kN

x

M

23.76 kNm

67 kNm

-72.5 kNm

1.64 m

Statically Indeterminate Beams

Superposition Method

P

L/2 L/2A

FA_x

FA_y

MA

Equilibrium conditions:

B

Compatibility Conditions

P

FB_y

C

F’B_y

C

Bε

(Example 12.23 p.681)

Statically Indeterminate Beams

Superposition Method

12-125 Determine the reactions at the support C. EI is constant. (p. 686)

PD

B

A C

L/2L/2

A C

F’B_y

PD

B

FB_y

1. Equilibrium conditions:

2. Superposition principle:

3. Compatibility conditions:

Symmetry condition:

Segment DB

Segment AC

Chapter 10 Buckling of Columns

Buckling of Columns

Buckling of Columns

Equilibrium conditions:

(second order differential equation)

Buckling of Columns

n=1 First mode of buckling

n=2 Second mode of buckling

n=3 Third mode of buckling

L

L

L/2 L/2

L/3 L/3L/3

Euler Load

Buckling of Columns

Equilibrium conditions:

Neutral Equilibrium:

Unstable Equilibrium:

Stable Equilibrium:

Buckling of Columns

13-7 The rod is made from a 25-mm diameter steel rod. Determine the critical load if theends are roller supported. E = 200 GPa, σY = 350 GPa (p. 707)

500 mm

PP1. Strength requirement:

2. Stability requirement: