The rigid bar AB, attached to two vertical rods as shown in
Summary
TLDRThe 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
🔍 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.
📏 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.
Mindmap
Keywords
💡Rigid Bar
💡Deflection
💡Vertical Rods
💡Horizontal Bar
💡Force P
💡Free Body Diagram
💡Reaction Force
💡Elastic Modulus (E)
💡Area (A)
💡Sigma (Σ)
💡Triangular Formula
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
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