The rigid bar AB, attached to two vertical rods as shown in

Summary

The video script discusses a structural engineering problem involving a rigid bar AB attached to vertical rods, subjected to a force P of 50 kilonewtons. The script guides through solving for the forces on aluminum and steel components, calculating their respective deflections, and determining the total deflection at point P. It uses principles of equilibrium and material properties to find the final deflection of 1.88 mm, providing a clear explanation for structural analysis.

Takeaways

• 📐 The problem involves a rigid bar AB attached to two vertical rods and a horizontal bar, with a force P applied to it.
• 🔍 The script describes a scenario where the deflection of both the aluminum and steel parts is the same, which is crucial for solving the problem.
• 📈 The force P is given as 50 kilonewtons, and the goal is to find the forces in the aluminum and steel parts.
• 📝 A free body diagram is drawn to visualize the forces acting on the system, including the applied force P and the reaction forces.
• ⚖️ The script applies the principle of equilibrium, stating that the sum of moments (rotational forces) around a point must equal zero.
• 📊 By setting up equations based on the moments, the force in the steel part (PST) is calculated to be 29.2 kilonewtons.
• 📐 The deflection of the steel part is then calculated using the formula involving force, length, area, and modulus of elasticity.
• 🔢 The deflection for the steel part is found to be 1.94 mm, and for the aluminum part, it is 1.78 mm.
• 📐 The script discusses the need to calculate the movement of point P, which involves understanding the geometry of the setup and the forces involved.
• 📐 The total deflection at point P (ΣB) is calculated by considering the deflection of the aluminum and steel parts and the angle formed by the forces.
• 🔑 The final answer for the total deflection at point P is given as 1.88 mm, which is the sum of the individual deflections and the additional movement due to the angle.

Q & A

What is the problem described in the transcript about?

The problem described in the transcript is about calculating the deflection of a rigid bar AB attached to two vertical rods and a horizontal bar, with a force P of 50 kN applied to it.

What are the two unknown forces that need to be determined in the problem?

The two unknown forces that need to be determined are the force in the aluminium (P_al) and the force in the steel (P_st).

What is the method used to find the unknown forces?

The method used to find the unknown forces involves drawing a free body diagram and applying the principle of equilibrium, specifically summation of forces (ΣF) equal to zero.

What is the value of the force P applied to the system?

The value of the force P applied to the system is 50 kN.

What are the distances used in the equilibrium equations?

The distances used in the equilibrium equations are 2.5 m and 3.5 m.

What is the calculated force in the steel (P_st)?

The calculated force in the steel (P_st) is 20 kN.

What are the parameters used to calculate the deflection of the steel and aluminium?

The parameters used to calculate the deflection are the force applied (P_st and P_al), the length of the material (L), the area of cross-section (A), and the modulus of elasticity (E).

What is the modulus of elasticity for steel and aluminium?

The modulus of elasticity for steel is 200 GPa, and for aluminium, it is 70 GPa.

What is the calculated deflection for steel and aluminium?

The calculated deflection for steel is 1.94 mm, and for aluminium, it is 1.78 mm.

How is the total deflection of the system calculated?

The total deflection of the system is calculated by considering the deflection of both the steel and aluminium components and the movement caused by the force P acting on the system.

What is the final answer for the total deflection of the system?

The final answer for the total deflection of the system is 1.88 mm.

Outlines

00:00

🔍 Structural Analysis of a Rigid Bar System

This paragraph discusses a structural engineering problem involving a rigid bar AB attached to two vertical rods and a horizontal bar. A force P, given as 50 kilonewtons, is applied, and the crucial aspect is the deflection or movement of the system. The speaker intends to solve for the forces in the aluminum and steel components, which are unknown. A free body diagram is suggested as the first step, followed by applying the summation of reaction force formula to find the forces. The paragraph concludes with calculating the force in the steel component (PST) as 20 kilonewtons using the given distance and force values.

05:01

📏 Calculation of Deflection in a Loaded Structure

The second paragraph continues the structural analysis by calculating the deflection of the steel and aluminum components under the applied force. The steel component's deflection is calculated using the modulus of elasticity (E), area (A), and the force applied (PST), resulting in a deflection of 1.94 mm. Similarly, the aluminum component's deflection is calculated with its respective values, yielding 1.78 mm. The paragraph then explains how to find the total deflection at point P by considering the movement caused by both the steel and aluminum components. The final deflection, denoted as Sigma B, is determined to be 1.88 mm after accounting for the angle and the respective movements in the structure.

Keywords

💡Rigid Bar

A 'rigid bar' is a structural element that is assumed to be inflexible and does not bend or twist under load. In the context of the video, the rigid bar AB is a key component of the structure being analyzed, attached to vertical rods and subjected to a force P, which results in deflection.

💡Deflection

Deflection refers to the displacement or bending of a structural element under an applied load. The video discusses the deflection of both the aluminum and steel components of the structure, which is crucial to understanding the structural response to the applied force P.

💡Vertical Rods

Vertical rods are the upright supports in the structure that the rigid bar AB is attached to. They play a significant role in the distribution of forces and the overall stability of the system described in the video.

💡Horizontal Bar

The 'horizontal bar' is another structural element that the rigid bar AB is connected to. It is mentioned in the script as part of the setup where the force P is applied, indicating its importance in the structural analysis.

💡Force P

In the script, 'Force P' is the applied force of 50 kilonewtons that causes the structure to deflect. It is a central element in the problem, as the entire analysis revolves around understanding the effects of this force on the structure.

💡Free Body Diagram

A 'free body diagram' is a graphical representation used to visualize the forces acting on an object in a system. The video mentions drawing a free body diagram as a first step in solving the problem, which helps in identifying and analyzing the forces at play.

💡Reaction Force

The 'reaction force' is the force exerted by a structure in response to an applied force, as per Newton's third law. The video script discusses calculating the reaction forces in the system to understand the equilibrium conditions.

💡Elastic Modulus (E)

The 'elastic modulus', often denoted as 'E', is a measure of a material's stiffness and is used in the formula to calculate deflection. In the script, the elastic modulus of steel is given as part of the data needed to find the deflection of the steel component.

💡Area (A)

The 'area' (A) of a cross-section of a structural element is important for calculating stress and deflection. In the context of the video, the area of the steel and aluminum components is used in the formulas to determine their respective deflections.

💡Sigma (Σ)

In the script, 'Sigma' (Σ) is used to denote the summation of values, particularly in the context of calculating the total deflection or movement of the point where force P is applied. It is a mathematical symbol used to represent the sum of a series of terms.

💡Triangular Formula

The 'triangular formula' mentioned in the script likely refers to trigonometric relationships used to calculate angles or other geometric properties in the structure. It is used to find the angle of rotation or the movement of the point where force P is applied.

Highlights

Problem involves a rigid bar AB attached to two vertical rods with a horizontal bar and a force P applied.

Deflection of both the bar and the applied force P is the same, indicating uniform movement.

Force P is given as 50 kilonewtons, and its effect on the system is crucial for the problem's solution.

A free body diagram is necessary to determine the unknown forces in the system.

Application of the summation of reaction force formula to find the unknown forces.

Calculation of the steel force (PST) using the given distance and force values.

Determination of the aluminium force (PAL) by considering the direction of forces.

Use of the formula for deflection involving force, length, area, and elastic modulus.

Calculation of the deflection for steel with given parameters.

Calculation of the deflection for aluminium using the force and dimensions provided.

Comparison of the deflections of steel and aluminium to understand the system's behavior.

Introduction of the concept of movement of point P and its significance in the problem.

Calculation of the total movement (Sigma P) by considering the deflections and distances.

Use of triangular formulas to calculate the angle and movement of the system.

Final calculation of the total deflection (Sigma B) combining all the calculated values.

Conclusion of the problem with the final deflection value of 1.88 mm.

Encouragement for viewers to ask questions in the comments for further clarification.

Invitation to subscribe to the channel for more problem-solving solutions.

Transcripts

this problem is saying that the rigid

bar AB attached to two vertical two

vertical this one vertical and this one

two vertical

rods shown in

figure

horizontally also is as attached to the

horizontal bar and P applied this p is

applied also P has given 50

kilon and the movement very crucial

point this movement you can say

deflection

deflection are same both are same

movement and deflection okay he they can

ask like this away and

movement Mo ask like this away and move

movement movement of Del A L that means

deflection of aluminium also they can

ask another that Sigma St deflection of

steel or movement of

Steel movement of Steel

okay

okay this we can ask in this question uh

now we'll solve this let at first we

know the five 50 kilon but we don't know

that this P this uh uh aluminium Force

we don't know or also we don't know the

steel Force okay that's why we have to

draw the free body diagram at first and

this is working the

PST and this's one is the P aluminium

okay and is acting on the

P now we'll apply we know this one is

the

50 kilon and distance 2.5 and this D is

3.5 M and then we have how we'll get

this we can if we apply

and

um

this summation of the reaction force

formula that that

means uh summation of

m a equal to zero obviously

anticlockwise is positive and p a l

equal into zero because another in this

for p p into 3.5 distance 3.5 distance

and is rot in in in this a this mean

negative and another is the plus PST for

is still distance is the total 2.5 and

3.5 this means 6 equal to 0 now we'll

get the PST because we know that one and

PST PST has a

29.2

Kon and another window know this no need

to

this this away we can uh we can sum we

know Direction according to direction we

know this one is positive you consider

this positive upward Direction positive

and downward Direction negative by by

this how we can

calculate and

then PST p P St plus p l minus P = to

Zer we know P we know P then easily

we'll get the

PST PST equal to 20 kilon okay now PS2

know

uh sorry p a l p a

l this one will be

p a l PS

PST uh PST yes p l yeah

now look we know this obviously we can

find out the easily deflection

ofel uh and PST into L l/

AE and BST 20

9.2 into PST this length is the 4 M and

area 300 mm square and E elastic

rigidity as 200 and then we'll get this

uh deflection for

uh for

STD

for S we get deflection of

1.94 another and for aluminum deflection

is the 20 into 3 / 500 into and 70 then

we'll get the

1.78 mm obviously this millim obviously

this

millim now we have to find out the

this for I have to

draw and this

sorry this

one okay this one if you consider this

deflection is coming for aluminium as

1.7

178 and this one is for 1

9

4 we know

1.78 of and P is acting over this we

have to uh p is acting on this we have

to calculate this movement this is the

movement of P okay P this mean Sigma P

we have to calculate this Sigma if we

draw this no this one but we know this

if you substract from biger to to

smaller value then we'll get the 1 7 mm

okay and this is the this

distance five and total is the six then

we can find out the a similar uh angle

triangular formula and this angle is

same for this a smaller a smaller

triangle

and also for this bigger triangle by

this we can calculate

3.5 / by a smaller just we identifying

this is for Sigma okay Sigma = to 6 /

by

1.7 and then Sigma =

to10 then because this one we know also

this one we know you have to to find out

the this I'm clear to you this don't

worry

just

I'm this one the sigma p and this p and

this one is the

sigma this Sigma deflection we know this

1 7 then total Sigma sorry sorry sorry

Sigma P we know this one this value we

know this value we have calculated

obviously this one we will get equal to

1.78

+10 that means

1.88 mm this is our deserv answer so

this deflection will be Sigma B = to

1.88 mm

I think it's clear to you though if you

have confusion you can ask in

comment or

also uh you you can subscribe my

channel important problem Sol solution

problem solving solution insh thank you

for watching