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Cornelis Kok
New Zealand
เข้าร่วมเมื่อ 23 ก.พ. 2016
A compolation of videos used in the New Zealand Diploma of Mechanical Engineering and Bachelor of Engineering Technology.
Example - Bolted connection with torsion and shear loads (Part 2 of 6)
A worked example going through a solution for the following problem.
A steel plate, which is fastened by a group of bolts, is subject to a 30kN load as shown below. All dimensions are in mm.
a) Calculate the centroid of the bolt group
b) Identify the bolt with the maximum load acting on it
c) Determine the maximum load acting on the bolt
d) Calculate the size of bolts required if the allowable shear stress in a bolt is 190MPa
A steel plate, which is fastened by a group of bolts, is subject to a 30kN load as shown below. All dimensions are in mm.
a) Calculate the centroid of the bolt group
b) Identify the bolt with the maximum load acting on it
c) Determine the maximum load acting on the bolt
d) Calculate the size of bolts required if the allowable shear stress in a bolt is 190MPa
มุมมอง: 16 911
วีดีโอ
Example - Bolted connection with torsion and shear loads (Part 4 of 6)
มุมมอง 12K5 ปีที่แล้ว
A worked example going through a solution for the following problem. A steel plate, which is fastened by a group of bolts, is subject to a 30kN load as shown below. All dimensions are in mm. a) Calculate the centroid of the bolt group b) Identify the bolt with the maximum load acting on it c) Determine the maximum load acting on the bolt d) Calculate the size of bolts required if the allowable ...
Example - Bolted connection with torsion and shear loads (Part 6 of 6)
มุมมอง 5K5 ปีที่แล้ว
A worked example going through a solution for the following problem. A steel plate, which is fastened by a group of bolts, is subject to a 30kN load as shown below. All dimensions are in mm. a) Calculate the centroid of the bolt group b) Identify the bolt with the maximum load acting on it c) Determine the maximum load acting on the bolt d) Calculate the size of bolts required if the allowable ...
Example - Bolted connection with torsion and shear loads (Part 3 of 6)
มุมมอง 12K5 ปีที่แล้ว
A worked example going through a solution for the following problem. A steel plate, which is fastened by a group of bolts, is subject to a 30kN load as shown below. All dimensions are in mm. a) Calculate the centroid of the bolt group b) Identify the bolt with the maximum load acting on it c) Determine the maximum load acting on the bolt d) Calculate the size of bolts required if the allowable ...
Example - Bolted connection with torsion and shear loads (Part 5 of 6)
มุมมอง 8K5 ปีที่แล้ว
A worked example going through a solution for the following problem. A steel plate, which is fastened by a group of bolts, is subject to a 30kN load as shown below. All dimensions are in mm. a) Calculate the centroid of the bolt group b) Identify the bolt with the maximum load acting on it c) Determine the maximum load acting on the bolt d) Calculate the size of bolts required if the allowable ...
Example - Bolted connection with torsion and shear loads (Part 1 of 6)
มุมมอง 21K5 ปีที่แล้ว
A worked example going through a solution for the following problem. A steel plate, which is fastened by a group of bolts, is subject to a 30kN load as shown below. All dimensions are in mm. a) Calculate the centroid of the bolt group b) Identify the bolt with the maximum load acting on it c) Determine the maximum load acting on the bolt d) Calculate the size of bolts required if the allowable ...
Polygon of Forces - Construction
มุมมอง 9K5 ปีที่แล้ว
This is part 3 (of 3) of a playlist related to the construction of a polygon of forces from a laboratory experiment.
Polygon of Forces - Laboratory
มุมมอง 4.8K5 ปีที่แล้ว
This is part 1 (of 3) of a playlist related to the construction of a polygon of forces from a laboratory experiment.
Polygon of Forces - Free Body Diagram
มุมมอง 4.8K5 ปีที่แล้ว
This is part 2 (of 3) of a playlist related to the construction of a polygon of forces from a laboratory experiment.
Block on an incline with acceleration
มุมมอง 2.4K6 ปีที่แล้ว
Video Solution fo the following question: For the 150kg block shown in the figure below, determine the force required (P) to push the block up the incline with an acceleration of 2.5m/s2. The kinetic friction coefficient is μk=0.3 between the block and the inclined plane.
Example - Max speed for car on Horizontal Track around constant radius curve (Part 2 of 2)
มุมมอง 3.4K6 ปีที่แล้ว
The sports car is travelling along a 30 banked road having a radius of curvature of ρ=150m. If the car does not change height, and the coefficient of static friction between the tires and the road is μS=0.2, determine the maximum safe speed so no slipping occurs. Neglect the size of the car.
Example - Max speed for car on Horizontal Track around constant radius curve (Part 1 of 2)
มุมมอง 18K6 ปีที่แล้ว
The sports car is travelling along a 30 banked road having a radius of curvature of ρ=150m. If the car does not change height, and the coefficient of static friction between the tires and the road is μS=0.2, determine the maximum safe speed so no slipping occurs. Neglect the size of the car.
Single Rivet, Double Shear - Example
มุมมอง 44K6 ปีที่แล้ว
A single rivet holds three sheets of steel together, and is loaded as shown below. If the maximum average shear stress allowed for the material is 125MN/m2, and a factor of safety of 3 is required by the design standards, find the minimum rivet diameter.
Method of Joints Example - Brief explanation - Part 1 of 2
มุมมอง 6366 ปีที่แล้ว
The following video is in response to a calculation error for an example in Hibbler - Engineering Mechanics - Statics
Method of Joints Example - Brief explanation - Part 2 of 2
มุมมอง 2206 ปีที่แล้ว
The following video is in response to a calculation error for an example in Hibbler - Engineering Mechanics - Statics
Two Connected Blocks on an Incline Part 3
มุมมอง 9487 ปีที่แล้ว
Two Connected Blocks on an Incline Part 3
Two Connected Blocks on an Incline Part 2
มุมมอง 1.7K7 ปีที่แล้ว
Two Connected Blocks on an Incline Part 2
Two Connected Blocks on an Incline Part 1
มุมมอง 4.2K7 ปีที่แล้ว
Two Connected Blocks on an Incline Part 1
Two Connected Blocks on an Incline Part 4
มุมมอง 7167 ปีที่แล้ว
Two Connected Blocks on an Incline Part 4
Energy Conservation - Moving object against a beam Example Part 3 of 3
มุมมอง 837 ปีที่แล้ว
Energy Conservation - Moving object against a beam Example Part 3 of 3
Energy Conservation - Moving object against a beam Example Part 2 of 3
มุมมอง 837 ปีที่แล้ว
Energy Conservation - Moving object against a beam Example Part 2 of 3
Energy Conservation - Moving object against a beam Example Part 1 of 3
มุมมอง 1707 ปีที่แล้ว
Energy Conservation - Moving object against a beam Example Part 1 of 3
Friction - Ladder against a wall part 2 of 2
มุมมอง 1.5K7 ปีที่แล้ว
Friction - Ladder against a wall part 2 of 2
Friction - Ladder against a wall part 1 of 2
มุมมอง 27K7 ปีที่แล้ว
Friction - Ladder against a wall part 1 of 2
Friction 04 Maximum Kinetic Friction Force
มุมมอง 9297 ปีที่แล้ว
Friction 04 Maximum Kinetic Friction Force
Friction 03 Maximum Static Friction Force
มุมมอง 2.3K7 ปีที่แล้ว
Friction 03 Maximum Static Friction Force
Friction 01 Introduction to Friction Lab Equipment
มุมมอง 9977 ปีที่แล้ว
Friction 01 Introduction to Friction Lab Equipment
Multiple stage constant acceleration equation question
มุมมอง 5557 ปีที่แล้ว
Multiple stage constant acceleration equation question
Thanks for the video. But while working I didn't see you taking into consideration the quadrant of the various functions (sin, cos) 🙄Like in the second quadrant cos is negative
Nice ☺️😊
Watching in 2024 thank you so much 🙏🇿🇲🇿🇲
That's one nice green marker....
Why do we have the center of the circle as 2 while we're given a diameter of 3 😢 Cause what I know is that the center of a circle is determined by the radius
clear,neat and understandable for all pupil thanks a lot.
why sin and why cos ?
Okay
5 cos 150 should be -5 cos 150 since it's on negative sign of x
Please note that cos 150 is negative, so there is no need for an additional negative sign.
Why is it 1/3 and 1/2 in terms of its distance?
The video was great but aren’t you supposed to subtract 360-18.5 since it’s in the fourth quadrant
Yes, that would be suitable. Both answers represent the same direction.
i dont get answer when using calculator
Check if your calculator is set to degrees (it might be set to radians or grad)
isnt the centroid of a triangle the 1/3rd and not the 2/3rds as you put?
he made a mistake,i notice that too
@@madzivhandiladecent6761 parallel to the Y axis
you do not consider the negative and positive signs
I have considered it by use of the sine and cosine functions. For example, you will notice that the x component for the 5N force will come up as negative.
What if the height is given
Why is the angle negative
Angles that are clockwise in the system are treated as negative angles and vice versa
sayup imohaa
Love it ❤️🫡
Sir for the Area of the circle, the radius in 1.5 not 3. And why did Sir divide the area of the circle by 4?
There is an equation for area of a circle using radius: pi * radius * radius And an equation for area of a circle using diameter: pi * diameter * diameter / 4 It is more common that the diameter of objects are given rather than the radius so the second formula is more useful. A google search will also provide you with the same formula as I have used. "Area of a circle using diameter"
This is the best method, hands down. Cheers
It's more than a help,, thanks for the great job
you might of just saved my T levels cheers boss
Hindiya me na bol skte h
aren't you supposed to subtract them because they are not going on the same direction??
Idk if this is too late but He already took into account the opposite directions when taking the angles from the x axis. You would only do the subtraction stuff if you left the angles the way they were given in the diagram.
Thanks sir for the refreshment because it’s been like six years now I studied this. If I am not confused I believe the angle for the resultant force should be 360°-18.5° since all the angles of the other forces were taken from the posive x-axis for the calculation
Thanks for the lesson way more easier than the way my physics teacher taught me
lifesaver, my maths lecturer only explained everything using 2 vectors but then gave us an assignment to find the resultant of 3 vectors so thank you!
How did the angle of the final answer became negative from positive 🙏
❤ appreciate it
why are they all having positive sign
Good lecture
Thank you Sir 🎉🎉🎉😮😮😮
Why do you choose negative angle when it is on the positive side of the horizontal line?
How did you calculate the square root of cos and sin cause I got a different answers
Me in 2024❤
Wow Very explained
So does this mean that in the x component we are suppose to use a cosine and in the y component we use sin
I'm watching in 2024❤
Same here
Same here
Same here
Me too got a test tomorrow 😅
Frr
Ok
Great vid !!!
what is the direction?
Pls why was the angle negative
How to report?
thank you.
How did the angle become a negative at the final answer
I am doing engineering level 3 and I joined in late when they already finished this topic. I am grateful that I finally have learnt from you. Thanks for the video again. Keep on posting more great stuff.
Bro still in high school😢 and we do this in Fm
this guy just save me, thanks !
I greatly appreciate. No wonder a day teaches for one to meet his destination.
Why didn't you use the first components?? Sorry am new to this
We could have ignored the first part. There are two approaches to breaking down the forces into components shown in this video. The first approach treats the vectors as triangles, where we find the size of the components, then add up the x components (if right, it would be added; and to the left, it would be negative). The second approach treats the vector as a phasor (try google this term and you should see a video of a vector that rotates, and that this rotating vector can be used to draw a sine wave (for y component) and cosine wave (for x component)
Shout out to those watching before physics externals