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3/15/2014 KAJIDAYA BAHAN: SHEAR AND MOMENT DIAGRAMS

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Wednesday, 6 July 2011

SHEAR AND MOMENT DIAGRAMS

MECHANICS - CASE STUDY

Introduction

Truck Bed Failure due to Loading

( 2003, Kurt Gramoll)

"Well, this is a new one for me" said the highway

patrolman when he was called to investigate the

accident. "I have never seen a semi snap in two due

to the load. I wonder why it failed near the front and

not in the middle?" His partner added, "I am sure the

investigation teamwill let us know. I wonder if the

truck was overloaded?"

What is known:

The trailer is 53 feet long (does not

include the t ractor).

The trailer gross weight (total including

both load and trailer) is 79,500 lb.

Assume weight is evenly distributed.

The truck bed is designed to withstand a

bending moment of 200 ft-kip (= 200,000

ft-lb).

Assume the truck bed can be modeled

as a simple supported beam with

supports at the left edge and at the

center of the back wheels.

Question

Assumed Location of Trailer Supports

Does the failure location agree with a basic beam

analysis? What is the value of the maximum

bending moment and does it exceed the design

moment of 200 kip-ft?

Approach

Construct a free body diagram.

Using static equilibrium equations,

determine the assumed support

reactions.

Determine how many sections are

needed and make a cut in each section.

Sum forces and moments at each cut in

terms of position.

Plot the resulting moment and shear

equations

MECHANICS - THEORY

Introduction

It is important to know how the shear forces and

FORMULA

GALLERY

SYNOPSIS

LECTURE

ANI MATI ON

CONTENT

2011 (29)

July(29)

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Multiple Beam Sections for

Different Loading Configurations

Multiple Beam Sections (or Segments)

One confusing aspect of developing moment and

shear diagrams is the need to have separate moment

and shear functions for each beam segment. This is

because a single function cannot model the moment

(or shear) change over a load or support (it is a

discontinuous function).

Each new beam section will have its own moment

and shear equations as a function of the location, x.

The diagram at the left gives various beams and

typical loadings.

After each section is cut, then a FBD is drawn for

either side of the beam, just like a single section

beam discussed above. Then the M and V is

determined for that cut and plotted.

Example: Multiple Beam Sections

Moment and shear diagrams are best understood by

examining an example. The simple support beam at

the left has a single point load between the supports.

The first step is to draw a FBD of the whole beam

and solve for the reactions.

MA= 0

(10 ft) RB- (6 ft)(120 lb) = 0

RB= 72 lb

Fy= 0

RA- 120 lb + 72 lb = 0

RA= 48 lb

Determine Beam Section

Section (1), to the left of the applied load, will have an

expression for the shear force and bending moment

that will differ from the section (2), to the right of the

applied load. Therefore, the sections must be

evaluated separately and each will have their own

moment and shear equations.

Section (1) (0 x 6 ft)

First, cut section (1) a distance x from the left side

and form a FBD, as shown. Remember, the

discarded right beam section needs to be replaced

by unknown an internal shear force and moment, V 1

an M1.

The left beam section must stay in static equilibrium.

Fy= 0

48 lb - V1= 0

V1= 48 lb

Mcut= 0

M1- x (48 lb) = 0

M1= 48x ft-lb

Section (2) (6 x 10 ft)

Now the next and last section can be cut. The left or


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.

section (1), the left side will be examined. Again,

form a FBD, as shown. The internal loads are

labeled, V2an M2to distinguish them from section

(1) shear and moment equations.

Again, applying the static equilibrium equations

gives,

Fy= 0

48 lb - 120 lb - V2= 0

V2= -72 lb

Mcut= 0

M2- x (48 lb) + (x - 6)(120 lb) = 0

M1= 720 - 72x ft-lb

Plot Shear and Moment Diagrams

The functions for V and M for both beam sections

can be plotted to give the shear and moment over the

length of the beam. The plots are given at the left.

The location for maximum and minimum shear force

and bending moment are easily found and evaluated.

Complex Distributed Load Example

Distributed loadingis one of the most complex

loading when constructing shear and moment

diagrams. This causes higher order polynomial

equations for the shear and moment equations.

Recall, distributed loadscan be converted to

equivalent forces which are easier to work with. Also,

complex, non-uniform distributed loads can be s plit

into simpler dist ributed loads and t reated separately.

An example is the best way to illustrate how to work

with non-uniform distributed loads. Take a simple

cantilever beam with a linear varying distributed load

as shown at the left. This example has only one

beam segment, so only one cut will be needed.

Cut the beam some distance x from the left. If the

right part of the cut beam is used, then the support

reactions at A do not need to be determined (this is a

unique situation).

The distributed load intensity, wcut, is a simple linear

relationship between A and B, or

wcut= 200 + (x/20)(500-200)

= 200 + 15x N/cm

The distributed load can be split into two parts, a

rectangular and triangular shape. The equivalent

loads, F1and F2of each shape are calculated as

F1= (20 - x) (200 + 15x)

= 4,000 + 100x - 15x2

F2= 0.5 (20 - x) [500- (200 + 15x)]

= 3,000 - 300x + 7.5x2

The internal moment and shear, M and V, can now

be determined using the equilibrium equations,

Fy= 0

V = (4,000 + 100x - 15x2) + (3,000 - 300x +

7.5x2)

V = 7,000 - 200x - 7.5x2 N

Mcut= 0

M = -[(1/2) (20-x)] (4,000 + 100x - 15x 2)

+ [(2/3) (20-x)] (3,000 - 300x + 7.5x2)

M = -80,000 + 7000x - 100x2- 2.5x3 N-cm

Both the shear and moment can now be plotted as a

function of position, x, to give the moment/shear

diagrams. Since the distributed load was changing
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near y, e s ear s a qua ra c equa on an e

moment is a cubic.

Moment and Shear Relationship

From the examples given in this section, it is

possible to s ee an interesting relationship between

the shear and the moment. The shear at a given

location is the slope of the moment function. In

mathematical terms, this is written as,

V = dM/dx

This relationship may be helpful in determining both

the moment and shear diagram without calculating

the actual function. Derivation of this relationship is

done in the Integration of Load Equation.

MECHANICS - CASE STUDY

Example Graphic

Beam Model with Loading

Introduction

The trailer is 53 ft long and carries 79,500 lb. Since

the load is assumed to be evenly distributed across

the trailer bed, the load intensity, w, is

w = (79,500 lb)/(53 ft) = 1,500 lb/ft = 1.5 kip/ft

The trailer is support by two sets of wheels which

can be modeled as simple pin joints as shown at

the left. Joint A is at the left edge and joint B is at

the center of the back wheels. Both of the support

joints are approximate.

Free Body Diagram with Reactions

Free Body Diagram

The basic trailer can be modeled with a free body

diagram to assist in finding the reaction forces.

There will be a reaction at joint A and B. These can

be found by using the static equilibrium equations,

MA= 0

39 RB- 26.5 (79,500) = 0

RB= 54,020 lb = 54.02 kipThe reaction at A can be found by summing forces,

Fy= 0

RA+ 54.02 - 79.5 = 0

RA= 25.48 kip

Beam Sections

Sections

Now that the reaction forces are known, the beam

can be cut into sections. This beam has one

continuous load and two supports. Since one of the

supports, the right one, is not at the beam edge,

this will cause a discontinuity in the moment and

shear diagrams. The beam will need to be analyzed

at two places, once in each of the sections shown in

the diagram.

Section 1)

Section 1 Cut

The first section is c ut and the left part of the cut

beam is used as shown in the diagram. The

unknown internal shear V1and moment M1are

applied at the cut edge to keep the beam sect ion in

equilibrium. These two loads can be determined

from the equilibrium equations, giving

Fy= 0

25.48 - 1.5x - V1= 0

V1= 25.48 - 1.5x 0 x 39 ft

Mcut= 0
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1+ x . x - . x =

M1= 25.48x - 0.75x2 0 x 39 ft

Section 2)

Section 2 Cut

The second cut is done between joint B and the

right edge of the beam. This time, the right beam

section will be analyzed. Again, the unknown

internal shear V2and moment M2are applied at the

cut edge. Note, that the direction of the internal

loads are opposite from section 1 c ut. This is

because the opposite beam section (the right side)

is used. This follows the sign conventionspecified in

the previous section. Applying the equilibrium

equations gives,

Fy= 0

V2- 1.5 (53 - x) = 0

V2= 79.5 - 1.5x 39 x 53 ft

Mcut= 0

-M2- [(53 - x)/2] [1.5 (53 - x)] = 0

M2= -2,107 + 79.5x - 0.75x2 39 x 53 ft

Plot S hear and Moment Equations

Shear and Moment Diagrams

Each of the two segments have different functions

for the shear and moment. These can be plotted

over each section to give a complete shear andmoment diagram. These are shown at the left.

With many shear and moment diagrams, a

maximum is needed. In this problem, the first beam

section increases and then decreases. To find the

location of the maximum, equate the first derivative

with respect to the locations, x, to zero. This gives,

dM1/dx = 0

25.48 - 1.5x = 0

x = 16.99 ft

The maximum moment at 16.99 ft is 216.4 kip-ft

which is a large moment. The moment is positive so

the beam will be bent downward. This moment does

exceed the maximum allowed of 200 kip-ft, so failure

would be expected.

Truck Failed Near the Location of

the Maximum Moment

Discussion

It is interesting to note that the maximum moment

from the analysis agrees well with the location

where the trailer failed

MECHANICS - EXAMPLE

Example 1

Example Graphic

A simple supported beam needs to support two

loads, a point force of 500 lb and a distributed load of

50 lb/ft as shown. Plot the shear and moment over

the length of the beam. What is the maximum shear

and moment?

In this example, there is a point load and a distributed

load. This will require the beam to be sectioned into

three segments. Each segment will have a separate

moment and shear equation.

Solution 1

The first step in analyzing any beam is to determine
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Newer Post Older PostHome

Posted by MOHD RAFIDI at 21:42

Free-Body Diagram

the reaction forces at the supports. Using the free

body diagram at the left, the moments can be

summed at point A to give,

MA= 0

12 RD- 9 (300) - 3 (500) = 0

RD= 350 lb

The reaction at A can be found by summing forces,

Fy= 0

RA- 500 - 300 + 350 = 0

RA= 450 lb

Beam Sections

Section 1 with Sign Convention

The beam needs to be split into three sections. Eachsection starts and ends at a load or support. After the

beam is organized into sections, then each section

can be cut and the internal shear and moment

determined.

Section 1

The first cut is between points A and B. The location

of the cut is not set, and thus labeled as x. For any

location x between x = 0 and x = 3 ft, the shear and

moment are given by

V = 450 0 x 3 ft

M = 450x 0 x 3 ft

Section 2

Section 2

The second cut is done between points B and C. This

beam section includes the 500 lb point force. The

section is kept in equilibrium by internal loads, V2

and M2. These two loads can be determined from the

equilibrium equations, giving

Fy= 0

450 - 500 - V2= 0

V2= -50 lb 3 x 6 ft

Mcut= 0

M2+ 500 (x - 3) - 450 x = 0

M2= 1,500 - 50x 3 x 6 ft

Section 3

Section 3

The third cut is done between points C and D (left

edge). This beam section is the most complex due to

the distributed that is cut. Only that part of the

distributed load that is on the section is used. The

width of the distributed load on the remaining beam

section is (x - 6) ft.

Just like the other two sections, the unknown internal

loads, V3and M3, can be determined from the

equilibrium equations, giving

Fy= 0

450 - 500 - (x - 6) 50 - V 3= 0

V3= 250 - 50x lb 6 x 9 ft

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