Welcome back, in this video we are going to solve problem 1-23 that is taken from chapter number 1 stress and book name is mechanics of material by R.C. Hibbler. The next segment is solve problem 1-22 for resultant internal loading acting on cross section passing through the handle arm at E and cross section of short link BC. So we will use the same data for problem 1-22 which we have solved in our previous video and now we will solve this problem 1-23.
So you can see 120 Newton load is applied on the handle. So we have to find the internal loading at cross section E and in link B. So let's start with the solutions.
First step is that if you remove this link at point B, so you will replace it with a reaction force and this reaction force that link applied at point B is equal to FBC. And when you remove this support, pin support at point A so you will be having reaction horizontal reaction force AX as well as vertical reaction force AY so we will find only FBC because we will then work on the left right portion right hand side of the problem and we will find is this FBC so far FBC will apply tuition of equilibrium that is sum of all moment about point A is equal to zero and taking the counterclockwise movement as positive. So you can see the load is applied like this and this FBC is vertical. So what we will do is that we will convert this load FBC into its component.
One is along this direction and other one is along this direction. So you can see that this angle is 60. So this angle will be 30. And if I extend this line, so these are the vertical vertical angle so this is 30 degree now this FBC will have two component one is along this direction and other is along this direction there or you can say this this component will pass through point A so this is FBC sine of 30 degree and this component is FBC cause of 30 degree this is the same FBC sign of 30 degree because these two reactions forces must pass through point B okay now sum of all moment about point a must be equal to zero so first load is 120 and perpendicular distance is this 500 and this is producing clockwise so it will be negative so I will write it first 120 into 500 that is clockwise so it will be negative the second moment is due to this load fbc into cos of 30 and perpendicular distance is this 50 and this is producing counterclockwise moment so it will be positive so i will write plus fbc into cos of 30 degree into 55 sorry uh 50 50 millimeter is equal to zero so from here when you calculate this FBC will be equal to 120 into 500 divided by 50 into cos of 30 degree and you will get this FBC will be equal to 1385.6 Newton or you can say 1.3738 56 kilo Newton so this is FC reaction force force in member BC okay now you can see if this is the member BC when you cut it over here so you can see the load that is taken by this member BC is FBC FBC so when you cut it you will be having a shear force you will be having a normal force clear and let's say this shear force is v this normal force is n and you will be having a movement m okay so we'll find this uh internal uh resultant loading in member or in short link BC so for that I will apply equation of equilibrium that sum of all forces along y direction must be equal to zero and upward force is taken as positive. So you can see this FBC which is this one 1385. minus this normal force is equal to zero so normal force in member in short link BC will be equal to one three eight five eight five point six Newton and that is 1.3856 kilo Newton so this is the normal force in member BC now for shear force we will apply equation of equilibrium that sum of all forces along X direction must be equal to 0 and force in this direction is taken as positive so you can see in this direction only ship force is shear force there is no other force so it means that V is equal to zero now we'll find moment and let this take point is point one so sum of all moment about point one is equal to zero taking the counterclockwise moment as positive so again at point one this force is passing this force is passing and there is only external moment over here so it means that moment will be also equal to zero and these are the answers for a resultant internet loading in short link DC. Now we will move toward this part and we will cut this handle at point E and we will draw the free body diagram.
So let me draw it first let this is the handle portion Okay, so at this point we have a load of 120 Newton 120 Newton this is point e so when you cut it so you will be having a normal force perpendicular to this face and this face is this is equal to ne you will be having a shear force along the surface of this this is VE and you will be having a moment which is ME okay the distance between this E and this this line is you can check the bow as well this is 100 millimeter and from this till this point this distance is 300 millimeter Now what we will do is that we will apply equation of equilibrium that sum of all forces along x dash is equal to zero. So what is x dash? So you can see this like this distance is taken as x dash and this coordinate is taken as y dash. We will resolve this.
forces into this X dash and y dash so first equation of equilibrium is this one and force taken in this direction is taken as positive so you can see the the NE which is acting in this direction is only force there is no other force so it means that NE will be equal to 0 so I will write NE is equal to 0 Now we will apply another equation of equilibrium that is sum of all forces along y direction must be equal to 0, y dash direction must be equal to 0 and force in this direction is taken as positive. So from here you can see one force along this y dash direction is this VE and the second force is this 120. So this is taken as positive and this will be taken as negative. Their sum must be equal to 0. So VE.
VE minus 120 Newton is equal to 0, so from here VE is equal to 120 Newton. And the last thing is moment about point E. So we'll apply equation of equilibrium that sum of all moments about point E is equal to 0 taking the counterclockwise moment as positive. So about point E one moment is this one which is counterclockwise so it will be positive and the second. Movement that is produced due to this 120 and perpendicular distance is 400 millimeter and this is producing clockwise so it will be negative and their sum must be equal to zero so i will write m e minus 120 into perpendicular distance 400 which in term of meter is 0.4 is equal to 0. so from here you will get moment about point e comes out to be 48 newton into meter So this NE, VE and ME is the resultant internal loadings at point E.
which is the answer of our this part so that was all about this problem 1-22 in which we have been asked to find out the internal loading in member bc which we have find out this normal force shear force and movement and internal loading at cross section at point e which is ne is equal to zero shear force is 120 newton and movement is 48 newton into meter i hope you have enjoyed this video and you have learned from it those who are new to my channel then subscribe it and don't forget to press the bell icon so that you can get notification about my latest video if you have any question you can ask me in comment section thank you for watching