Determine internal resultant loading | 1-22 | stress | shear force | Mechanics of materials rc hibb

Summary

This educational video tackles problem 1-22 from 'Mechanics of Material' by RC Hibler, focusing on stress analysis. It guides viewers through calculating reactive forces at a pin and link BC of a metal structure under a 120 Newton force. The video then explains how to determine the internal loading on a cross-section at point D. Using equilibrium equations and geometrical analysis, the presenter solves for various forces and moments, offering a clear understanding of mechanics in materials.

Takeaways

• 🔧 The problem is from 'Mechanics of Material' by RC Hibler, specifically chapter 1 on stress.
• 📐 A 120 Newton force is applied to a metal structure punch, and the task is to determine the reactive forces at pin A and link BC.
• 🔍 To find the reaction force at pin A, the structure is analyzed using equilibrium equations, considering moments about point A.
• 📏 The reaction force at pin A is calculated by resolving the forces into horizontal (ax) and vertical (ay) components.
• 📐 The force in link BC (FBC) is found to be 13856 Newtons or approximately 13.86 kN using the equilibrium equations.
• 🔄 The vertical reaction force at point A (ay) is determined to be 14.89 Newtons or 1.49 kN, considering the vertical components of the forces.
• 🔄 The horizontal reaction force at point A (ax) is calculated to be 60 Newtons using the equilibrium of forces in the horizontal direction.
• 📐 The resultant reaction force at pin A is calculated using the Pythagorean theorem, yielding approximately 1.49 kN.
• 🔍 For the internal loading at cross-section D, the structure is sectioned, and a free body diagram is used to determine the normal force (ND), shear force (VD), and bending moment (MD).
• 📏 The internal loading at point D is found with ND being 120 Newtons, VD being 0 Newtons, and MD being 36 Newton-meters, indicating the forces and moments acting on the structure at that point.

Q & A

What is the force applied on the metal sturred punch handle?

The force applied on the handle is 120 Newtons.

What are the two components of the reaction force at point B when link BC is removed?

The two components of the reaction force at point B are the horizontal component (FBCx) and the vertical component (FBCy).

How is the angle of 30° used in determining the components of FBC?

The angle of 30° is used to calculate the horizontal and vertical components of the reaction force FBC using trigonometric functions, specifically cosine and sine.

What is the magnitude of the reaction force FBC in the link BC?

The magnitude of the reaction force FBC is approximately 13856 Newtons or 13.85 Kilo-Newtons when rounded.

What are the equations of equilibrium used to find the reaction forces at point A?

The equations of equilibrium used are the sum of all forces along the y-direction and the sum of all forces along the x-direction, which must equal zero.

What is the vertical reaction force (ay) at point A?

The vertical reaction force (ay) at point A is approximately 14.89 Newtons or 1.49 Kilo-Newtons.

What is the horizontal reaction force (ax) at point A?

The horizontal reaction force (ax) at point A is 60 Newtons.

How is the resultant reaction force at point A calculated?

The resultant reaction force at point A is calculated using the Pythagorean theorem by combining the horizontal and vertical components (ax and ay).

What is the internal loading acting on the cross-section at point D?

The internal loading at point D includes the normal force (ND), shear force (VD), and bending moment (MD).

What are the values of ND, VD, and MD at point D?

The values are ND = 120 Newtons, VD = 0 Newtons, and MD = 36 Newton-meters.

How is the bending moment (MD) at point D calculated?

The bending moment (MD) at point D is calculated by considering the moment due to the 120 Newton force at a perpendicular distance of 0.3 meters from point D.

Outlines

00:00

🔧 Problem Introduction and Reaction Force Calculation

This paragraph introduces the problem from 'Mechanics of Material' by RC Hibler, specifically problem 1-22. The scenario involves a metal sturred punch subjected to a 120 Newton force on the handle. The task is to determine the reactive force at Pin A and link BC. The solution begins by conceptually removing link BC and pin support at Point A to identify the reaction forces FBC and the components ax and ay. The equilibrium equations are applied to find FBC, which involves calculating moments about Point A, considering the 120 Newton load and its perpendicular distance to the point of interest. The components of FBC are resolved using trigonometric functions, leading to the calculation of FBC as approximately 138.56 Newtons or 1.39 Kilo-Newtons.

05:03

📏 Calculation of Reaction Forces at Pin A

The second paragraph continues the solution by determining the vertical and horizontal components of the reaction force at Pin A, denoted as ay and ax respectively. The vertical component ay is calculated using the equilibrium of forces in the y-direction, considering the vertical component of FBC and the 120 Newton load's vertical component. The result is ay as approximately 9.56 Newtons or 1.5 Kilo-Newtons. The horizontal component ax is found by applying the equilibrium of forces in the x-direction, yielding ax as 60 Newtons. The paragraph concludes with the calculation of the resultant reaction force at Pin A using the Pythagorean theorem, which is approximately 49.1 Newtons or 1.49 Kilo-Newtons.

10:06

🛠 Internal Loading at Cross-Section D

The final paragraph addresses the second part of the problem, which is to determine the internal loading acting on the cross-section at Point D. The solution involves cutting the beam at D and analyzing the free body diagram, which includes forces and moments acting on the section. The equilibrium equations are used to calculate the normal force ND, shear force VD, and bending moment MD at Point D. ND is found to be equal to the applied load of 120 Newtons, VD is determined to be zero due to the lack of forces in the y-direction, and MD is calculated based on the moment arm and the applied load, resulting in a moment of 36 Newton-meters. The paragraph concludes with the summary of the internal loadings at Point D, which are ND, VD, and MD.

Keywords

💡Mechanics of Materials

Mechanics of Materials is a branch of engineering that studies the behavior of materials under various mechanical loads. It is crucial for designing structures and components that can withstand different types of forces. In the video, the book 'Mechanics of Material by RC Hibler' is referenced, indicating that the problem being solved is related to this field. The video's theme revolves around applying the principles of mechanics of materials to solve a specific engineering problem.

💡Stress

Stress is a physical quantity that describes the internal resistance of a material to deformation when a force is applied. It is calculated as the force applied per unit area and is a fundamental concept in mechanics of materials. In the video, the problem involves a metal structure subjected to force, and the calculation of stress would be essential to understand how the material responds to the applied load.

💡Punch

In the context of the video, a punch refers to a mechanical device or tool used to apply force, often in the form of a punch press. The punch is the point of application for the 120 Newton force mentioned in the problem, which is used to analyze the reaction forces and internal loadings in the structure.

💡Reaction Force

Reaction force is the force exerted by a support in response to an applied force. It is a consequence of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In the video, the calculation of reaction forces at pin A and in link BC is a key part of the problem-solving process, demonstrating how these forces balance the applied force on the punch.

💡Equilibrium

Equilibrium refers to a state in which the sum of all forces and moments acting on a body is zero, resulting in no acceleration. The video uses the principle of equilibrium to derive equations that help calculate the reaction forces and moments. This concept is central to the analysis of the mechanical system described in the problem.

💡Moment

A moment, in mechanics, is a measure of the turning effect of a force about a point or an axis. It is calculated as the product of the force and the perpendicular distance from the line of action of the force to the point or axis. In the video, moments are calculated to analyze the rotational effects of forces on the structure, particularly in the context of equilibrium conditions.

💡Component

In the context of forces, a component refers to the individual parts of a force vector when it is resolved into orthogonal (perpendicular) directions, typically horizontal (x) and vertical (y). The video discusses how to break down forces into their x and y components to simplify the analysis and calculation of forces and moments.

💡Internal Resultant Loading

Internal resultant loading refers to the collective effect of all the internal forces and moments acting on a cross-section of a material. In the video, determining the internal loading at a specific cross-section (point D) is part of the analysis to understand how the material behaves under the applied load.

💡Cross-Section

A cross-section is a virtual slice through a structure that allows for the analysis of the internal forces and moments at that point. In the video, the cross-section passing through the handle arm at point D is analyzed to determine the internal loading, which is crucial for understanding the structural integrity.

💡Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that states the relationship between the sides of a right-angled triangle. In the video, the theorem is applied to calculate the resultant force at pin A by combining the horizontal and vertical components of the reaction forces.

Highlights

Problem 1-22 from the book 'Mechanics of Material' by RC Hibler involves a metal structure punch subjected to a force.

The force applied on the handle is 120 Newtons.

The task is to determine the reactive force at Pin A and in the short link BC.

The internal resultant loading on the cross-section at Point D through the handle arm must also be found.

By removing link BC, a reaction force FBC at point B is introduced.

Removing the pin support at Point A results in horizontal and vertical reaction forces ax and ay.

Equilibrium equations are used to calculate the reaction forces.

The moment about Point A is calculated considering the 120 Newton load and the perpendicular distance.

The FBC force has x and y components, which are resolved using trigonometric functions.

The equation FBCx * 50 = 0 is used to find the x-component of FBC.

The magnitude of FBC is calculated to be approximately 13856 Newtons or 13.85 Kilo-Newtons.

The vertical component of the reaction force at Point A (ay) is found using the equilibrium of forces in the y-direction.

The horizontal component of the reaction force at Point A (ax) is determined to be 60 Newtons.

The resultant reactive force at Pin A is calculated using the Pythagorean theorem.

The internal loading at Point D is determined by cutting the beam and drawing a free body diagram.

The normal force (ND) at Point D is found to be equal to the applied load of 120 Newtons.

The shear force (VD) at Point D is calculated to be zero.

The bending moment (MD) at Point D is determined using the equilibrium of moments about Point D.

The internal loading at Point D consists of ND, VD, and MD.

The video concludes with a summary of the problem's solution and an invitation for viewers to subscribe and engage.

Transcripts

welcome back in this video we are going

to solve problem 1-22 that is taken from

chapter number one stress and the book

name is mechanics of material by RC

hibler so statement is the metal sturred

punch is subjected to force of 120

newton on the handle determine the

magnitude of reactive Force at Pin a and

in short link BC also determine the

internal resultant loading acting on the

cross-section passing through handle arm

at D so you can can see 120 newton load

is applied on handle and we have to find

the reaction force at the pin as well as

in link BC and then at Point D we have

to find the internal loading so let's

start with the solution so first step is

that if you remove this uh Link at Point

uh if you remove this link BC so there

will be a reaction force at point B let

this uh Force force is

FBC and if you remove this pin support

at Point a so you will be having a

reaction force horizontal which is ax

and there will be a vertical reaction

force which is represented as ay now you

can find the uh this reaction force by

using equation of equilibrium so first

equation of equilibrium is that sum of

all for sum of all movements about point

a is equal to zero and taking the

counterclockwise mement as positive so

from here you can see that this this

angle is 60° clear this angle is 60° so

this will be

30° and if I draw line like this so This

angle will be also 30° now FBC have two

uh two component one is this x component

and other is this y component clear so

we can also represent this component

like this this will be

FBC y component and this will be

FBC X component okay so about point a

the one moment is produced due to this

120 newton load and perpendicular

distance is 500 mm and that is 0.5 M and

this will produce clockwise rotation so

it will be negative so I will write it

first

120 into

500 and that is negative the second

moment is produced due to this component

of fbcx component there and

perpendicular distance is you can see

that is 50 mm and this is producing

clockwise movement so it will be

positive the vertical component is

passing through a is not producing M so

I will write plus FB C into x

* 50 is equal to 0 so - 120 into

500 and FBC FB CX is equal to FBC into

COS of 30° FBC into COS of 30° * by 50

is equal to0 so from here you will get

this FBC will be equal to

13856 Newton or you can say

138 we will get if you round it

1.39 Kil Newton so this is the value of

FBC so in first uh part we have find out

the for magnitude of force in short link

BC which is FBC now we'll move toward

finding ax and a y for that we will

apply equation of equili that sum of all

forces along y direction must be equal

to zero and upward force is taken as

positive so you can see vertically we

have this one force a y clear the second

one is this Force FB BC and we will take

the uh vertical component of this so let

me first Define the vertical component

if I draw a vertical line over here so

you can see this angle is 30°

how this line is perpend okay you can

see that the angle between these two

line is 30° and this line is

perpendicular to this clear and this

vertical line is perpendicular to this

line so the angle between these two will

be also 30° so now this 120 will have

two component one is this

component and other one is this

component so this component is equal to

120 120 into s of 30° and this component

will be 120 into COS of

30° so now you can see this is the

vertical Force this is the vertical

force and this is the vertical force and

their sum must be equal to zero so I

will write minus FBC FBC is

13856 plus a y which is upward minus

this component which is 120 into

COS of 30° their sum must be equal to

zero so from here when you calculate it

you will get a y will be equal to 14 8

9.56 Newton R you can see

1.5 Kil

newon so this is the vertical component

of reaction force at Point a now we'll F

find this ax by using another equation

of equilibrium that sum of all forces

along X direction must be equal to zero

and force in this direction is taken as

positive so you can see we have one

force which is ax and other one is this

component there some must be equal to 0

so I will write - a

x + 1 2 0 into s of 30 is De is equal to

0 so from here here you will get ax will

be equal to 60

Newton so this is the answer of our

first part in which we have been asked

to determine the magnitude of reactive

Force at Pin a and short link BC now the

resultant of this horizontal and

vertical force will be can be obtained

by using Pythagoras Theorem and let this

is reaction force a r f a

so we can find this fa a reactive Force

at Pin a will be equal to ax² + a y²

under the root so put the value ax is 60

s + a y is 14 8

9.56 squ Under The Root so here when you

calculate it will be

491 Newton or you can see that Force at

Point a will be equal to

1.49 Kil

newon okay now in second part we have

been asked to determine the internal

loading acting on cross-section passing

through handle arm at D so for that we

will cut the beam over here at D and we

will draw the pre body diagram so pre

body diagram will be like this one so

let this is

the handle

portion okay and then you

have let me correct

it and then at Point D so D is sorry

this one D is this one so when you cut

it you will get it like

this okay so here you can see you

have 120 newton

force that is acting perpendicular to

this

surface 120

nton and at Point D when you cut it you

will be having a share Force which is

this is point D which is equal to VD you

will be having a normal force which is

equal to n d and you will be having a

moment which is equal to MD so how you

will get the distance between this and

the point of application of 120 load is

given as 300

mm now you can see we have to change the

coordinates so what we will te is that

we will take this x coordinate in this

direction because here we have normal

load and the share force will be in this

direction this is Y dash so we will

apply equation of equilibrium that first

equation of equilibrium is that sum of

all forces along x d must be equal to

zero and force in this direction is

taken as positive so you can see one

force is this one which is n d and the

second Force which is this one

their sum must be equal to zero so I

will write n d - 120 is equal to0 so n d

will be equal

to 120

newton now we'll find the

second force that is VD so for that

we'll apply equation of equum that sum

of all forces along y Das direction must

must be equal to zero and force in this

direction is taken as positive so you

can see along this we have one force is

VD and there is no other Force so it

means that only VD so VD is equal to

zero okay we'll find this moment MD by

using equation of equilibrium that sum

of all Ms about Point D is equal to zero

and taking the counterclockwise mement

as positive so about this point d one

moment is this MD which is

counterclockwise so it will be positive

the second M will be at Point D will be

due to this 120 and perpendicular

distance is this one which in meter is

0.3 and this is producing clockwise so

it will be negative so I will write MD -

120 into perpendicular distance is 0.3 m

is equal to zero so when you calculate

it you will get M about Point D comes

out out to be 36 Newton into

M so these are ND VD and MD are internal

loading at Point D and that is the

answer of our second part and this was

all about this problem 1-22 I hope you

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