SFD and BMD - Problem 1 - Part 1 - Shear Force and Bending Moment Diagram - Strength of Materials

Summary

This script details the process of analyzing an overhanging beam's structural behavior under various point loads. It begins with calculating support reactions at points A and B using equilibrium conditions, resulting in RA = 16.71 kN and RB = 32.09 kN. The script then proceeds to determine the shear force diagram (SFD) and bending moment diagram (BMD), crucial for understanding the beam's stress distribution. The SFD reveals varying forces along the beam, while the BMD is essential for identifying the point of contraflexure, a key structural feature. The explanation is technical, targeting an audience with an engineering background.

Takeaways

• 📚 The task involves analyzing a structural engineering problem concerning an overhanging beam with specific support points and loads.
• 🔍 The primary goal is to determine the support reactions at points A and B, which are essential for understanding the beam's stability.
• ⚖️ The equilibrium condition is applied by summing vertical forces (FY) to zero, considering upward forces as positive and downward forces as negative.
• 📐 Moments are calculated about point A, with clockwise moments being positive and anti-clockwise moments negative, to find the reaction force RB.
• 🔢 Through calculations, the reaction force RB is determined to be 32.29 kilo Newtons, which is a critical value for the beam's support.
• 📉 The reaction force RA is then found by subtracting RB from the total vertical force, resulting in RA being 16.71 kilo Newtons.
• 📈 Shear force diagrams (SFD) are to be drawn to visualize how the forces vary along the length of the beam.
• 📍 The sign convention for shear force calculations is established, with upward forces to the left and downward forces to the right considered positive.
• 📉 Shear force at point C is calculated considering the forces and reactions, resulting in a value that remains constant up to point A.
• 🔄 The shear force changes at different sections, such as at point D and B, reflecting the impact of the loads and reactions on the beam's integrity.
• 📝 The final step involves identifying the point of contraflexure, which is a location along the beam where the curvature changes sign, indicating a change in the bending moment.

Q & A

What is the main objective of the script?

The main objective of the script is to solve a structural engineering problem involving an overhanging beam, which includes determining support reactions at points A and B, drawing the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD), and finding the point of contraflexure.

What are the given point loads on the overhanging beam?

The given point loads on the overhanging beam are 10 kN at 2 meters from A, 24 kN at 4 meters from A, and 15 kN at 10 meters from A.

What are the equilibrium conditions used to find the support reactions?

The equilibrium conditions used are the summation of vertical forces (ΣFY) and moments about a point (ΣMA), which are set to zero according to the principles of static equilibrium.

What is the calculated reaction at support A (RA)?

The calculated reaction at support A (RA) is 16.71 kN.

What is the calculated reaction at support B (RB)?

The calculated reaction at support B (RB) is 32.29 kN.

How are the moments about point A calculated?

The moments about point A are calculated by considering the clockwise moments as positive and the anti-clockwise moments as negative, and summing these moments to zero.

What is the sign convention for shear force calculations?

The sign convention for shear force calculations is that upward forces to the left and downward forces to the right are positive, while downward forces to the left and upward forces to the right are negative.

What is the shear force at point C?

The shear force at point C is -10 kN, considering the downward force to the left of the section as negative.

How does the shear force change from point A to point D?

The shear force at point A is +6.71 kN, and it remains constant up to the left of point D. To the right of point D, the shear force becomes -17.29 kN after considering the reactions and loads.

What is the shear force at point B?

The shear force at point B is +15 kN, which remains constant up to point E.

What is the next step after calculating the support reactions and shear forces?

The next step after calculating the support reactions and shear forces is to draw the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD), and then find the point of contraflexure.

Outlines

00:00

🏗️ Structural Analysis of Overhanging Beam

The first paragraph introduces a structural engineering problem involving an overhanging beam with specific support points and point loads. The task is to determine the support reactions at points A and B, draw the Shear Force Diagram (SFD) and Bending Moment Diagram (BMD), and identify the point of contraflexure. The solution begins with the equilibrium condition, calculating the vertical forces RA and RB, treating upward forces as positive and downward forces as negative. The calculation leads to an equation involving RA and RB, which are then solved using moment equilibrium around point A, yielding RA as 16.71 kN and RB as 32.29 kN. This section concludes with the determination of the support reactions.

05:01

📊 Shear Force Calculations for Beam Structure

The second paragraph continues the structural analysis by detailing the process of calculating the shear forces along the beam. The sign convention for shear force is established, with upward forces to the left and downward forces to the right considered positive, and vice versa. The calculation starts by examining the section close to point C, determining the shear force at various points along the beam, including at points A, D, B, and E. The shear force values are calculated by considering the forces acting to the left and right of a chosen section, leading to a constant shear force between certain points. The final values determined are -6.71 kN at point A, -17.29 kN at point D, and +15 kN at points B and E, concluding the shear force calculations for the beam.

Keywords

💡Overhanging Beam

An overhanging beam is a structural element that extends beyond its main supports, often used in construction to provide additional space or reach. In the video's context, the overhanging beam is the primary subject of the analysis, where the speaker discusses determining support reactions and other structural properties.

💡Support Reactions

Support reactions refer to the forces exerted by the supports on a structure in response to the loads applied to it. In the script, the speaker calculates the reactions at points A and B to ensure equilibrium, which is a fundamental concept in structural analysis.

💡Equilibrium Condition

The equilibrium condition is a principle stating that the sum of all forces and moments acting on a body must be zero for the body to be in equilibrium. The video script uses this principle to calculate support reactions, ensuring that the vertical forces (FY) sum up to zero.

💡Shear Force Diagram (SFD)

A shear force diagram is a graphical representation that shows the variation of shear force along the length of a beam or other structural element. In the video, the SFD is calculated to understand the distribution of forces and to find the point of contraflexure.

💡Bending Moment Diagram (BMD)

A bending moment diagram illustrates the bending moments in a beam, which are crucial for understanding how the beam will deform under load. The script mentions calculating the BMD as part of the structural analysis process.

💡Point of Contraflexure

The point of contraflexure is the location along a beam where the bending moment changes sign, indicating a change from compression to tension or vice versa. The script aims to find this point as part of the structural analysis.

💡Kilo Newton

Kilo Newton is a unit of force in the International System of Units (SI), equivalent to one thousand Newtons. In the script, the loads on the beam are measured in kilo Newtons, which helps in calculating the support reactions and other structural properties.

💡Moments

In the context of the video, moments refer to the torque or turning effect of a force about a particular point. The speaker calculates moments about point A to find the support reactions, considering the direction and distance of the forces from the point of interest.

💡Sign Convention

Sign convention is a method of assigning positive or negative signs to forces and moments based on their direction relative to a chosen reference. In the script, the speaker uses a sign convention to determine the direction of shear forces and bending moments, which is essential for accurate calculations.

💡Shear Force

Shear force is the force that tends to cause sliding or deformation parallel to the cross-section of a structure. The script describes calculating shear forces at various points along the beam to understand the internal forces acting on it.

💡Bending Moment

Bending moment is the internal force couple within a structure that tends to bend it. The script discusses calculating bending moments to analyze the beam's behavior under the applied loads and to create the BMD.

Highlights

The task is to determine support reactions at points A and B for an overhanging beam.

The beam has point loads of 10, 24, and 15 kilo Newtons.

Equilibrium conditions are applied to calculate support reactions RA and RB.

Vertical forces are considered with upward forces as positive and downward as negative.

Summation of vertical forces (ΣFY) is set to zero for equilibrium.

Moments about point A are calculated with clockwise moments as positive and anti-clockwise as negative.

Moment calculations involve distances from point A and the direction of forces.

RA and RB values are determined using equilibrium equations.

RA is calculated to be 16.71 kilo Newtons.

RB is determined to be 32.29 kilo Newtons.

Shear force diagram (SFD) and bending moment diagram (BMD) are to be drawn next.

Sign convention for SFD is established with upward forces to the left as positive.

Shear force at point C is calculated considering the forces to the left of the section.

Shear force at point A is determined by considering the forces on the left of section A.

Shear force at point B is calculated in two ways, yielding the same result of +15 kilo Newtons.

Shear force at point E is constant and equal to +15 kilo Newtons.

The process involves calculating the point of contraflexure, though not detailed in the transcript.

Transcripts

Let us take this first question for an

overhanging beam shown in Figure

determine support reactions at A and B

draw SFD in BMD and find the point of

contraflexure now this is the question

which we have in front of us let us try

to write the data first here I will draw

the beam which is given

now this is the question which we have

in front of us where we have an

overhanging beam which is supported at

Point A at point B then we have point

load of 10 kilo Newton 2400 10 and 15

kilo Newton acting on this beam for this

diagram we have to find out the support

reactions the first question is we have

to calculate reaction at a and reaction

at B that is I have to determine our a

value and our B value the next part is

we have to calculate s FD and bmd and

after that we have to find the point of

contraflexure so this is the question

which we have I will write the solution

for this I'll say that for the

equilibrium condition of the beam that

is my step number one is calculation of

support reactions here I will say that

for the equilibrium condition first

summation of FY is equal to zero where I

will treat upward forces positive and

downward force is negative so therefore

I have the sum of vertical forces as RA

is upward RB is upward so both are

positive that is RA plus RB 10 24 and 15

they are downwards so negative minus 10

minus 24 and minus 15 is equal to 0

therefore we have RA plus RB is equal to

if I add these terms it will be 10 plus

24 plus 15 which which is 49 here I have

negative 49 if I shift it on to the

other sides it becomes positive

so here I have my first equation after

this I will be taking moments at Point a

so I will say that summation of moments

about point a is equal to 0 where

clockwise moments are positive and

anti-clockwise moments are negative so

therefore if I am taking moment at Point

a I start from R be the moment is

anti-clockwise and distance is 7 meter

between point a and B so it is minus RB

into 7 next here I have 15 kilo Newton

which is at a distance of 10 meters from

point A and it is in clockwise direction

so I will take it as plus 15 into 10

then I have 24 kilo Newton which is at 4

meters from a and it is in clockwise

direction so plus 24 into 4 next I have

10 its distance is 2 meters from point A

and if I take the moment about point a

it is an anti-clockwise direction so

minus 10 into 2 therefore if I calculate

all terms here I will be getting RB as

it comes out to be 32 point 2 9 kilo

Newton and after this I will say that

put RB is equal to 32 point 2 9 kilo

Newton

in equation 1 so therefore I have RA it

was plus RB so plus 32 point 2 9 is

equal to 49

therefore RA is equal to 49 minus 32

point two nine

therefore RA value it comes out to be

sixteen point seven one kilo Newton if I

subtract these values so here I have

found out the support reactions and this

completes the first part that is I have

got the first answer RA RB values now I

will be writing the values here

ra is 16.7 1 RB is 32 point O nine so

here I have found the support reactions

the next part is I have to calculate

shear force so my step number two is SF

calculations and for that I will draw

the sign convention first upward force

to left and downward force to the right

are positive downward force to left and

upward force to right they are negative

so therefore I will start calculating

the value of shear force first I will be

taking the section at you can say very

close to Point C this is my section I

will see that first SF at Point C

is equal to here I have a section and to

the left of it that is downward force so

downward force to the left is taken as

negative value is minus 10 kilo Newton

next this shear force will remain

constant between C to a up to left of a

now when I take the section to the right

of a and if I look to the left of this

section I have minus 10 and plus sixteen

point seven one so SF at Point a minus

10 plus sixteen point seven one that

comes out to be plus six point seven one

kilo Newton now this value of shear

force at Point a will remain constant

from A to D up to the left of D if I

take the section to the right of D then

I have minus ten plus sixteen point

seven one minus 24 this value of shear

force it comes out to be minus seventeen

point two nine kilo Newton next SF at

point B is equal to if I take the

section to the right of B here I can

calculate it in two different ways

either I can see the left of section or

right of the section the answer will be

same so here if I take left I will have

minus ten plus sixteen point seven one

up hold

- 24 downward +32 0.09 upward so I have

the value as plus 15 kilo Newton and

this value will remain constant up to

point E so I say that SF at Point E is

equal to plus 15 kilo Newton so these

are the SF calculations