2-Lectures LEC 02 Load and Stress Analysis

Chapter Outline

4.1 Equilibrium and Free-Body Diagram

4.2 Shear Force and Bending Moments in Beams

4.3 Singularity Functions (Omitted)

4.4 Stress

4.5 Cartesian stress Components

4.6 Mohr's Circle for Plane Stress

4.7 General Three-Dimensional Stress

4.8 Elastic Strain

4.9 Uniformly Distributed Stresses

4.10 Normal Stresses for Beams in Bending

4.11 Beams with Asymmetrical Sections (Omitted)

4.12 Shear Stresses for Beams in Bending

4.13 Torsion

4.14 Stress Concentration

4.15 Stress in Pressurized Cylinders

4.16 Stresses in Rotating Rings (Omitted)

4.17 Press and Shrink Fits

Chapter 4

Load and Stress Analyis

A. Bazoune

LEC-02

Equilibrium and Free-Body Diagram

Shear Force and Bending Moments in Beams

Chapter 4: Load and Stress Analysis

ME 307 MACHINE DESIGN I

3/10 LEC-02 Dr. A. Aziz Bazoune

4.1 Equilibrium and Free-Body Diagram

Equilibrium System: used to denote any isolated part or portion of a machine or structure. A system may consist of

• a particle, or several particles,

• a part of rigid body, an entire rigid body, or several rigid bodies. Equilibrium: A system is said to be in equilibrium if it is motionless or has a constant velocity, i.e.; zero acceleration. The phrase static equilibrium is also used to imply that the system is at rest. For equilibrium, the forces and moments acting on the system balance such that

( )Vectorial Representationx y zF F F+ + =∑ ∑ ∑ ∑F = i j k 0� � ��

(4-1)

or

( )Sc

0

0

0

alar Representationx

y

z

F

F

F

=

=

=

∑∑∑

and

( )Vectorial Representationx y zM M M+ + =∑ ∑ ∑ ∑M= i j k 0� � ��

(4-2)

or

( )Sc

0

0

0

alar Representationx

y

z

M

M

M

=

=

=

∑∑∑

Free-Body Diagram (FBD) Free Body Diagram (FBD) is a sketch that shows a body of interest isolated from all interacting bodies. Once the body of interest is selected, the forces and moments exerted by all other bodies on the one being considered must be determined and shown in the diagram, see Figure 4.1

1F�

2F�

3F�

4F�

Figure 4-1 The isolation of a subsystem

Using FBD for force analysis serves the following important points:




• The diagram establishes the directions of reference axes, provides a place to record the dimensions of the subsystems and the magnitudes and directions of the known forces and helps in assuming the directions of unknown forces.

• The diagram simplifies your thinking because it provides a place to store one thought while proceeding to the next.

• The diagram provides a means of communicating your thoughts clearly and unambiguously to other people.

• Careful and complete construction of the diagram clarifies fuzzy thinking by bringing out various points that are not always apparent in the statement or in the geometry of the total problem. Thus, the diagram aids in understanding all facets of the problem.

• The diagram helps in recording progress in the solution and in illustrating the method used.

• The diagram allows others to follow your reasoning, showing all forces.

CONSTRUCTING A FREE BODY DIAGRAM (FBD) STEP 1: Decide which body or combination of bodies is to be shown on the FBD. STEP 2: Prepare a drawing or sketch of the outline of this FBD. STEP 3: Carefully trace around the boundary of the free body and indentify all the forces

exerted by contacting or attracting bodies were that were removed during the isolation process.

STEP 3: Choose a set of coordinate axes to be used in solving the problem and indicate these directions on the FBD. Place on the diagram any dimensions required for solution of the problem.

EXAMPLE 4-1 Draw a free body diagram for the beam shown in Fig. 4-2a.

Solution Referring to Figure 4-2 (b):

• Two concentrated forces

1P and

2P are applied to the

beam.

• The weight of the beam is

represented by the force W , which has a line of action that passes through the center of

gravity G of the beam.

Figure 4-2 (a)

Figure 4-2 (b)

• The beam is supported at the left end with a smooth pin and bracket and at the right end with a roller.

• The reaction of the left support is represented by the forces xA and yA .

• The reaction of the roller is represented by the force yB , which acts normal to

the surface of the beam.




EXAMPLE 4-2 A cylinder is supported on a smooth inclined surface by a two-bar frame as shown in Figure 4-3a. Assume that the

cylinder has a weight W and that the two bars have negligible weight. Draw a free body diagram for a) The cylinder b) The two-bar frame c) The pin at C.

Solution a) The FBD for the cylinder is shown in

Fig. 4-3(b). The weight W of the cylinder acts through the center of

gravity G . The forces 1

N and 2

N

act normal to the smooth surfaces at the points of contact.

b) The FBD for the two-bar frame is shown in Fig. 4-3(c). The action of the smooth pin and bracket supports at points A and C are represented by the

forces xA and yA and xC and yC ,

respectively. Notice that the pin forces at B are internal and do not appear on the FBD shown in Fig. 4-3(c).

c) Since bar BC is a link , the resultant

cF of forces xC and yC must have a

line of action along the axis of the link. As a result, the FBD for pin C can be drawn as shown in Fig.4-3 d.

Figure 4-3(a)

Figure 4-3(b)

Figure 4-3(c)

Figure 4-3(d)

EXAMPLE 4-3 (Example 4-1 in the Textbook) Figure 4-4(a) shows a simplified rendition of a gear reducer where the input and

output shafts AB and CD are rotating at constant speed iω and oω , respectively.

The input and output torques (torsional moments) are 240 lbf iniT = i and oT ,

respectively. The shafts are supported in the housing by bearings at A, B, C and D.

The pitch radii of gears 1G and

2G are

10.75 inr = and

21.5 inr = ,

respectively. Draw the FBD of each member and determine the net reaction forces and moments at all points.




Figure 4-4(a) Gear Reducer

Figure 4-4(b) Gear Box

Figure 4-4(c) Input shaft

Figure 4-4(b) Output shaft

Solution Simplifying assumptions:

1. Gears 1G and

2G are spur gears with a standard pressure angle 20φ = �

.

2. The bearings are self aligning and the shafts can be considered to be simply supported.

3. The weight of each member is negligible. 4. Friction is negligible. 5. The mounting bolts at E, F, H and I are of the same size.

The separate FBD of the members are shown in Figs 4-4 b-d. The force transmitted

between the spur gears is not tangential but at the pressure angle φ . Thus,

tanN F φ= (1)

Input Shaft AB




( )0 0.75 240 0 320 lbfxM F F= ⇒ − = ⇒ =∑ (2)

and the normal force is

tan 320tan20 116.5 lbfN F φ= = =� (3)

0 320 lbfy Ay ByF R R F= ⇒ + = =∑ (4)

0 116.5 lbfz Az BzF R R N= ⇒ + = =∑ (5)

( ) ( )0 2.5 1.5 0 0.6 lbf=192 lbfz Ay AyM R F R F= ⇒ − = ⇒ =∑ (6)

( ) ( )0 1.5 1.0 0 1.5y Bz Az Az Bz

M R R R R= ⇒ − = ⇒ =∑ (7)

Substitution of Eq. (6) into Eq. (4) gives 128 lbfByR = . Similarly, substitution

of Eq. (7) into Eq. (5) gives 46.6 lbfBzR = and 69.9 lbfAzR = .

Output Shaft CD Following the same procedure,

0 320 lbfy Dy CyF R R F= ⇒ + = =∑ (8)

0 116.5 lbfz Dz CzF R R N= ⇒ + = =∑ (9)

( ) ( )0 2.5 1.5 0 0.6z Cy Dy Cy DyM R R R R= ⇒ − = ⇒ =∑ (10)

( ) ( )0 1.0 1.5 0 1.5y Cz Dz Cz DzM R R R R= ⇒ − = ⇒ =∑ (11)

Substitution of Eq. (10) into Eq. (8) gives 192 lbfCyR = , therefore

320 192 128 lbfDyR = − = . Similarly, substitution of Eq. (11) into Eq. (9) gives

46.6 lbfDzR = , therefore 69.9 lbfCzR = .

The output moment is

( )240 320 0.75 480 lbf inoT = + = i

Notice that in Fig.4-4(b) the net force from bearing reactions is zero whereas the net moment about the x-axis is

( ) ( ) ( ) ( )1 2 1 2 1 2

2.25(192) 2.25(128) 720 lbf in

cy Dy cy DyT r r R r r R r r R R

T

= + + + = + +

+= = i




This value is the same as 240 480 720 lbf ini oT T+ = + = i as shown in Fig. 4-

4(a). (Remember

Gear Box

The reaction forces ER , FR , HR , and IR from the mounting bolts cannot be

determined from the equilibrium equations as there are too many unknowns. Only

three equations are available 0y z xF F M= = =∑ ∑ ∑ . In case you were

wondering about assumption 5, here is where we will use it. The gear box tends to

rotate about the x-axis because of pure torsional moment of 720 lbf ini . The bolt

forces must provide an equal but opposite torsional moment. The center of rotation relative to the bolts lies at the center of the centroid of the bolt cross-sectional area. Thus if the bolt areas are equal: the center of rotation is at the center of the

four bolts, a distance of ( ) ( )2 2

4 2 5 2 3.2015+ = in from each bolt; the

bolt forces are equal ( )E F H IR R R R R= = = = , and each bolt force is

perpendicular to the line from the bolt to the center of rotation. This gives a net torque from the four bolts of

( ) 74 23 0.2 lb i015 f nR = i .

Thus, 56.22 lbfE F H IR R R R= = = =

4.2 Shear Force and Bending Moments in Beams

Figure 4-5a shows a beam supported by reactions 1

R and 2

R and loaded by the concentrated

forces 1

F , 2

F , and 3

F . If the beam is cut at some section located at 1

x x= and the left

hand portion is removed as a free body, an internal shear force V and bending moment M must act on the cut surface to ensure equilibrium.

• The shear force is obtained by summing the forces on the isolated sections.

• The bending moment is the sum of the moments of the forces to the left of the section taken about an axis through the isolated section.

The sign conventions used for bending moment and shear force are shown in Figure 4-6.




Cut

V

M

Figure 4-5 FBD of simply supported beam with V and M shown in positive direction

Figure 4-6 Sign conventions for bending and shear.

Shear force and bending moment are related by

d

d

MV

x= (4-3)

Sometimes the bending is caused by a distributed load ( )q x , as shown in Figure 4-7; ( )q x is

called the load intensity with units of force per unit length and is positive in the positive y direction. It can be shown that differentiating Eq. (4-3) results in

2

2

d d

d d

V Mq

x x= = (4-4)




Figure 4-7 Distributed load on beam.

Integrating Eqs (4-3) and (4-4) result in

d dB B

A A

V x

B AV x

V q x V V= = −∫ ∫ (4-5)

which states that the change in shear force from A to B is equal to the area of the loading diagram between

Ax and Bx .

Similarly

d dB B

A A

M x

B AM x

M V x M M= = −∫ ∫ (4-6)

which states that the change in moment from A to B is equal to the area of the shear force diagram between

Ax and Bx .

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2-Lectures LEC 02 Load and Stress Analysis