Jointed RodsHeavy Rod frame works Day 04(Past Paper Qns)

Ajouté : 2026-05-22

[00:00:00]Anyone a few minutes of time?

[00:00:09]Okay, take one minute more.

[00:00:25]Question number 15. Now this is 20191.

[00:00:29]Two uniform routes AB and BC each of length 2 A jointed smoothly at B. The rot AB is of weight w and uh BC is 2 W.

[00:00:40]The ND inch is smoothly. Now they have given the two rots AB and BC makes angles alpha and beta with vertical like this. Okay.

[00:00:56]And again this rod makes an angle beta right beta and this is alpha.

[00:01:08]Okay.

[00:01:10]And again uh here at C the force has applied perpendicular to the rot. Now the amount is W / 2 and the weight of BC is 2 W.

[00:01:27]Now this is 2 W. Can understand right? Show that tan beta= 5 / right. Now this is B joint capital B.

[00:01:39]Now this is A smooth and B smooth.

[00:01:46]Now that first one we need to find what is beta right B that without thinking okay component of the reaction at B you can prove that beta= 5 / 6 simply taking moments about B for BC B can smooth joint okay but is not being rotated therefore sum of moments would be equal to zero for the road BC for the road capital BC I will take moments about B. When you consider moments about B, they have given the length is 2 A, right? A= B= 2 A. A= B= 2 A.

[00:02:28]Now when you take moments about B W 2 * 2 A W / 2 * 2 A minus right B 2 W * A sin beta 2 W * A sin beta equal 0 W to W A to A cancel you will get that sin beta equal half right sin beta= half therefore beta= 5 / 6 acute right beta= 5 / 6. Now we have taken what is beta is right. So that B take equal 56 and find the horizontal and vertical component of the reaction at the joint capital B under the root BC right exerted from the root A forces right now these are the reaction at the joint capital B now you know that this is 2 W down this is W / 2 that we found that this is beta now if I draw in a vertical line like This was 5.

[00:03:43]Therefore, this is 5. This is 3 upward can understand when you consider the vertical equator BC to XB should be in this manner directions right in this direction and here when you consider the joint the equibrium of the pin at B methane YB paril you can find what are the components actually that they are reaction right no find the horizontal and vertical component of the reaction at joint B okay for BC 4 BC You can resolve forces upward. When you resolve forces upward for VC, YB - 2 W - 2 W and again plus W / 2 30. Therefore, this is 30. Then this should be 60.

[00:05:05]Plus W / 2 cos 60 = 0. I'll make the subject YB. YB equal 2 W - W / 4 2 W - W / 4 mean 2 into 4 8 that's 7 W over 4 now this comes 7 W / 4 my vertical component at the reaction B vertical component vertical component at capital B now I will try to find what is XB to find XB you can resolve forces horizontally 4 BC. Okay.

[00:05:45]4 BC. When you resolve forces horizontally, you will get that W / 2 may 30 30 this is 60. W /2 sin 60 W / 2 sin 60 - XB equals 0. You can make subject XB sin 60 roo<unk>3 / 2 <unk>3 W / 4. Right?

[00:06:06]You'll get that XB= <unk>3 W / 4 right horizontal component at B horizontal component component at B. Right?

[00:06:29]Okay. Also show that the tan alpha is <unk>3 over 9. How can you find tan alpha now?

[00:06:36]know right what is XB and YB the weight of the road AB is W okay weight of the road AB is W now we can consider moments about A for the road AB only can understand then we know what are the forces acting at B on AB now we can take moments about A now this is very similar to framework the two rods right have in connected that's it now see for AB for AB you can take moments about Amay when you take moments about A XB time what was XB we have already taken see XB <unk>3 W / 4 <unk>3 W / 4 * 2 A sin alpha 2 A sin alpha right XB B is in this direction that YB and W both in the other way minus YB that YB was 7 W / 4 7 W / 4 * 2 A sin alpha 2 A sin alpha minus W * A sin alpha equ= 0. Now uh this should be uh <unk>3 w* 2 a cos alpha right the perpend okay 2 a cos alpha 2 a cos alpha 2 a cos alpha hurry a to a cancel w to w to w cancel now you can take like terms together now here you know that <unk>3 / 2 * cos alpha equal When you take DC the other way 7 / 2 + 1 how 7 / 2 + 1 will be what? 9 /2 right okay that comes 9 /2 now what's the answer is for here 9 /2 * sin alpha 2 to2 will be cancel tan alpha= tan alpha= <unk>3 / 9 okay that we have taken answer who got this one the answer right now this is 20191 Okay, how was the paper? That's easy. Very easy right now. The first two papers were little easy actually. But next we convert product standard.

[00:09:57]Question of 14.

[00:10:09]Question number 14. Who got the answer?

[00:10:12]Okay, great.

[00:10:20]Tala. Four equal uniform rods each of each of way W1 are smoothly jointed at their ends to form the rhombus A B C D.

[00:10:31]Midpoints of B, C and C D are connected by a light rod such that B A D equal to theta. Each of the joints B and D carries equal loads weight W2. The system hand symmetrical from the joint capital A is in equibrium in a vertical plane with the light rods horizontal so that the thrust of the light rod is this much this is the rhombus right. Okay. Now the four rods are equal in length. Now this corction is very similar to corrections we did okay regarding the square right.

[00:11:15]The only thing is instead of 45, theta has come. That's it. Right? And two additional weights are there at the two ends.

[00:11:26]Right?

[00:11:28]Now the two weights have been attached at B and A, B, B and D. Now this is A B. Now this is the joint capital C and now this is D. The midpoints of uh BC and CD have been connected. Hurry up.

[00:11:54]You know that the system tries to go down. Therefore, there should be a thrust.

[00:12:06]weights have been attached at B and D.

[00:12:10]Now what are the weights? W2 W2 W2 and now this is W2.

[00:12:18]The each rod the weight is W1 W1 and again this is W1 and now this comes W1 like that.

[00:12:36]Now the here right now this X component they find out is T right now we have to show that right now this S component probably should be in which direction since the joint is symmetric about C since the joint is symmetric about C the vertical line okay joint C truem we know that y goes zero right y now in which direction x should be now if you are considering mments about d for the ro cd you can see that this w rotate in this way since this is a thrust rotating this way right now this is x and therefore this should be x like that symmetric I have taken 2 y in the same direction.

[00:13:36]Right? Now the first step is we can say y equals 0 and can you know the y and i can of it as y y y y y y y y y y y y y y y y y y y y y z if you are considering the equibrium of the pin at C right okay the vertical equibrium of the okay now we can draw the forces for pin at C pin at C if it's like this now forces acting at pin at C will be like is y this x is here and x.

[00:14:14]Now when you consider the vertical equin that y should be equal to zcra that joint is symmetrical about the vertical line. Therefore y should be equal to z. Now you know that easily we can take y=0. Okay y= 0. Now when y=0 that we need to find what is t right anyway they don't ask what is xing okay trust of the properties what can we do we can take moments about d for cdt no a right okay that's when b aa right theta that because this is rhombus right Okay. Who got the answer again?

[00:15:07]Okay. Right. Now see for CD for CD you can take moments about D you can take moments about D. Now when you take moments about D for CD W * W * A sin theta right have they given us the length? No.

[00:15:26]Right. Now let let A be equal to A.

[00:15:32]Right? W * A sin theta W * A sin theta now W1 I think this is W1 W1 and W1 W1 * A sin theta plus T * A cos theta plus T * A cos theta minus X * 2 A cos theta - X * 2 A cos theta equals Z. Anyhow that we need to make the subject t I will take these things into one side and divide by cos theta equation w1 w1 tan theta w1 tan theta equ= - t1 + 2x + 2x take this equation as one.

[00:16:35]Okay.

[00:16:37]T1 minus P1.

[00:16:40]Right. I I will take this equation as one. And again we can take moments about a for the rod for the for two rod A and DC about four A comma DC I will consider moments about A when you consider moments about A for two rods A D and DC W1 * A sin theta on the perpend New distance a sin theta may a sin theta w1 * a sin theta plus again w1 * a sin theta okay and again okay okay in this way that capital x time you know that 2 a cos theta 2 a cos theta w2 and then plus w2 plus w2 * perpendicular distance w2 right 2 a sin theta 2 a sin theta plus capital x * the perpendicular distance to capital x. You can see 2 a cos theta 2 a cos theta 4 a cos theta 4 a cos theta capital x * 4 a cos theta minus t * - t * the perpendicular distance to t is a cos theta 2 a cos theta 3 a cos theta 3 a cos theta equal zero now what can we do is that you can cancel a to a to a to a and divide by cos theta again. So when you divide by cos theta you can see that 2 W1 tan theta plus 2 W2 tan theta 2 W1 + 2 W2 * tan theta equal take these things in the other way when you take these things in the other way okay that we have divided by cos theta when you take it in the other way - 4x - 4x + 3t this is equation two and that we need to make subject But we need to make subject t. No, but we can multiply this by two and that when you multiply this by two that 4x to 4x will be cancel from 1 and 2 from 1 and 2. See what's happening is now I'm multiplying this by two then 2 w1 tan theta plus 2 w1 tan theta that comes 4 w1 tan theta. Okay, that 4 W1 + 2 W2 * tan theta equ= right now here - 2t + 3t that comes t this one to this one will be cancel equal t is that the answer yeah right now the t equal 2 is common w1 sorry 2 w1 + w2 * tan theta maybe right answer again right this is 2016, right? The only thing is okay.

[00:19:51]Easily you can remove from C and fine.

[00:19:54]Right. Okay.

[00:21:52]Okay. Right.

[00:21:56]May now this is question number 13 the 2015 15 level. You can see that now this is actually a symmetrical system. Right?

[00:22:07]Now this is symmetrical about this line.

[00:22:09]Now this is a and uh a bi b c d e like this. This is pq like trot the main system. You can understand how this tries to go down. Therefore there will be a thrust in yes t a thrust. Now this is the that a and a e there are two equal rods. Okay equal in length 2.

[00:22:37]Now this part actually that P Q B C D E okay 2 L make a 2 L like this. Okay, right that because L L like that. Now the weights are 2 W and again this is 2 W. Now you know that the weight of this one W and weight of this W right W like that.

[00:23:14]Now the first they ask to find reaction at B.

[00:23:19]Okay.

[00:23:21]The framework is freely suspended from cap. Okay. Write down the equation sufficient determine the horizontal and vertical component of the reaction at joint capital B and the thrust T in the like. Hence find the reaction on the rod AB at the joint capital B like that.

[00:23:37]Bye-bye man. Now right now easily we can find uh component of the reaction at B actually.

[00:23:46]Can anyone of you tell the y component should be in upward direction? No downward direction probably in upward direction. Now okay.

[00:23:57]Okay. Now if you think that there are two ways to think the one way is what?

[00:24:00]When you think the equibrium of the part E, DC and CB E D and CB this is symmetrical. Now this is symmetric. Now okay symmetric line this is symmetric line.

[00:24:15]This is symmetric line.

[00:24:18]symmetric. You know that E D CB this part is in equilibrium that because of these weights and these forces since this is symmetric the mayometer in Y and even this is Y upward. Okay. Now the vertical equilibrium of these three rods okay because of www and these two y now the easily you can take okay you can resolve forces for these three upward and then you can find y what's the other way is what's the other method is finding y one method is what okay that consider you can take okay you can resolve forces for E D C and CB vertically upward sorry otherwise you can take moments about E for E D and C about E Right. Yum. Up and Y down. Where should be X? X.

[00:25:42]Okay.

[00:25:47]W.

[00:25:59]Right. You know that the distance is at 1 and 1 2 1.

[00:26:05]Okay. 3 W.

[00:26:27]Okay. We can write equations to find what is X and Y easily if you taking moments about E for the three roles E D C and CB you will get what is Y for E D for E D comma DC and CB and CB I will take moments about A. When you consider moments about a 30 right now this one is 30 therefore this is 60 they have given this length L and 2 L okay L and 2 L Y * L cos 60 L + L cos 60 y * L cos 60 + L + L cos 60 - W time minus W * the perpendicular distance to this W is L + L cos 60 + L / 2 cos 60 can understand L + L cos 60 + L / 2 cos 60 minus W * again the perpendicular distance this one L cos 60 + L / 2 L cos 60 + L / 2 minus again W time the perpendicular distance this one is L cos 60 right minus W * L cos 60 okay L sorry L /2 cos 60 now see W * L /2 cos 60 equals Z okay now you know that okay L2 L2 L2 L2 L will be cancel right now y * you know half + 1 and again plus half you can take these things with the The way W time take all together W time. Now see 1 +/ + 1 / 4.

[00:28:30]Right? Now here 1 +/ + 1 / 4. Now here again uh that comes one right + 1 and again + 1 oh no half right half + half right half + half and again plus 1 / 4 right + 1 / 4. What will happen? Oh here actually that half half that comes one and again one no wait that's this comes at then one way and again 1 / 4 1 / 4/ again 1 3 W okay now this is 2y 2 I = 3 W that y = 3 W / 2 y= 3 W now even you are going to resolve forces for the tree rods a d cb upward then y = 3 w / 2 unknown right now we can find what is x's to find x you can take mments about c only for the roc see only for the roc we can take moments about cand for bc about c moment Right? Now for BC when you take moments about C in this sense that capital X time this is 30 that capital X time L cos 30 capital X time L cos 30 X * L cos 30 capital X * L cos 30 in this way plus W * W * L / 2 cos 60 perpendicular distance at L / 2 cos 60 minus Y * - Y * Y V is 3 W / 2 - 3 W / 2 * perpendicular distance L cos 60 L cos 60 equals Z. Now you can cancel L2 to L2 to L cos 3 <unk>3 /2 <unk>3x / 2 equal when you take these things the other way 3 W over 4 minus W / 4 what will happen 3 W / 4 - W / 4 mean 2 W / 4 that comes W / 2 these two is cancel X= W / <unk>3 right now this is the vertical component these are the answers You got okay unknown? So that tal to find t. Now what can we do? You can take moments about a for a par we have resolved from a right about a moment for the ab right for a we can take moments about a for a I'm taking moments about a moment Right.

[00:31:47]When you take moments about a t * c 2 and you know that this angle is 30 t * l cos 30 t * l cos 30 h in this way now all others are in the other way right capital x y and 2 w - 2 w * l cos 60 L cos 60 - 2 W * L cos 60 - Y * 2 L cos 60 What was Y - 3 W / 2 * 2 L cos 60 2 L cos 60 - X * X * 2 L cos 30 what was X W / <unk>3 - W / <unk>3 * 12 L cos 30 2 cos 30 equ= 0. Now that L2 L to L will be cancel. You can make subject T T * cos 30 <unk>3 / 2 equal W * you can take these things in the other way. Elbows cancel. Now the w time 2 into cos 60 that comes one no one plus now the here l cancel 2 to 2 cancel and cos 60/2 then 3 / 2 + 3 / 2 plus when you take these things in the other way cos 3 <unk>3 / 2 then <unk>3 / 2 then 2 to 2 cancel roo<unk>3 to roo<unk>3 cancel again + one maybe right now here you will get That t = may 1 + 1 2 2 into 2 4 + 1 + 3 7 7 / 2 22 2 cancel 7 W / <unk>3 t= 7 W / <unk>3 we have taken the answer right okay that's really easy this is 2015 who got this answer Right.

[00:35:55]Question number nine.

[00:36:01]Right. Now this is 2011.

[00:36:07]Two uniform rots AB and BC are equal in length. The weights AB weight of AB is 2 W and the weight of BC is W. Smooth the hinge at B and the midpoint of the roots are connected by light is elastic. The system is stand in equilibrium in a vertical plane with A and C on a smooth horizontal table.

[00:36:29]A CN smooth horizontal table.

[00:36:37]A and C table right the midpoints have been attached tension that because of the two rows tries to expand like this A we need to find what's the reaction at B I think components at B and C system a A that because of weights not equal length equal weight equal symmetric. This is RC and this is R E.

[00:37:11]Now the weights are weights of AB is 2 W and the weight of U BC is W 2 W you know.

[00:37:24]We need to find what are the reactions are. See abc is 2 theta mega theta mega theta the system is stand in equilibrium if above that the tension of this is find the magnitude of the reaction at b and the angle it makes with horizontal right they don't ask what are the reaction at a and c we can Take moments about a for the entire system.

[00:38:01]When you consider the two roads as a single body internal forces are the forces acting at the joint capital on BCory.

[00:38:19]See for the system let a equal 2 a let= 2 a right for the system for the system I will take moments about a when you take moments about a moment about a for the entire system when you take moments about Therefore the system RC time RC time the perpendicular distance RC theta the 2 A cos theta 2 A cos theta that comes 4 A cos theta - W * - W * the perpendicular distance of this W A cos uh 4 A sin theta right now this is 4 A sin theta 4 A sin theta minus W * A sin theta 2 A sin sin theta.

[00:39:20]Now this comes 3 a sin theta minus 2 w * - 2 w * a sin theta equal z right then now this is a a a cancel sin theta to sin theta will be cancel rc equal you will get that 3 + 1 that uh 5 w / 4 rc= 5 W / 4. This is reaction at C. Reaction at C reaction at capital C.

[00:40:00]Okay.

[00:40:04]3 W sin theta 2 W * A sin theta.

[00:40:10]H R= 5 W 4. We need to find that so that the tension of the string. Okay.

[00:40:21]P we can take moments about B for BC. RC for BC for BC. We can take moments about B.

[00:40:33]When you take moments about B for BC, right?

[00:40:39]Right.

[00:40:45]T * A cos theta T * A cos theta minus sorry plus W plus W * the perpendicular distance to W is A sin theta plus W * A sin theta minus 5 W / 4 time perpend C 2 A sin theta 2 A sin theta equ= 0. Now you can say that when you take this one the other way that two 2 5 W / 2. Now the here that 5 W / 2 minus W that comes 3 W / 2 TNR TNR 3 W / 2 tan theta tension in the string tension in the string.

[00:41:40]tension in this string.

[00:41:44]Right? Find the magnitude of the reaction at B and the angle it make.

[00:41:50]Now when you consider the BC now the RC was 5 W 5 W 4 is more than W down can understand right these are the component of the reaction B XB YB B made now further out BC for BC I will resolve in this direction when you resolve in this direction XB minus P = Z then XB = T 3 W / 2 tan theta and again uh I will resolve in this direction when you resolve in this direction RC that comes 5 W / 4 - W - YB = Z now the IB equal W / 4. Now we have taken what are the component of the reaction at B. Now we can find what's the reaction at B. Reaction at B mean the resultant of these two reaction at B.

[00:43:10]Reaction at B reaction B specifically reaction at B on BC A.

[00:43:20]But if they asking find the reaction at B on BC BC right hand right hand side if they asking okay reaction at B on BC we need to take the resultant of these two reaction at B on A we need to take the resultant of these two take that XB and Y like this see XB B and Y X Y right this is RB RB= XB ^ 2 + YB ^ 2 is square root what was XB XB was 3 W /2 that's 9 9 W² / 4 time tan is theta plus y is w² / 4.

[00:44:30]Now you can say 1 + 9 tan² theta square<unk> / 2 * w right what is alpha now we need to find what's the direction is alpha equ= tan inverse yb / xb yb / xb what was yb yb was uh W 4 WW cancel two cancel that comes uh six right okay yb over uh xb no no y / xb right n 23 that theta / 2 theta / that y that goes right tan inverse theta / 6 this is what alpha y always been Sorry.

[00:45:53]W 4h 16 divide by four divide by four right 3 W.

[00:46:14]Yes. Yes. Yes.

[00:46:19]W / 4 ^ 2 and this comes 4 9 into 4 that comes 36 1 + 36 tan² theta / 4 right Take fun.

[00:48:13]Okay, very good.

[00:48:22]You can say that alpha alpha All right.

[00:48:37]System reaction.

[00:49:01]Right. A A A A B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B B C and uh C D Now this is horizontal force P right lambda W and uh 2 lambda W.

[00:49:18]Again uh they ask what are the horizontal vertical component of the reaction at B. Therefore we need to resolve from B.

[00:49:27]Now this is lambda W and this is W.

[00:49:30]Right? Now this W these are alpha alpha and again alpha alpha alpha alpha like this A B C D we can take mments about C for C D and you can find what is P in terms of lambda what is lambda first right and the horizontal and vertical component see for CD for CD I will take moments about C then when you take moments about C have they given what's the length is this A equal to A that A= 2 A when you take moments about C for CD P * 2 A cos alpha P * 2 A cos alpha minus 2 2 lambda W time A sin alpha equals Z now a 2 A cancel what you will get for B P= lambda W tan alpha lambda W tan alpha we have taken what's the value of P is P and again you can take moments about B for B C and C D then no mments coming from P Right?

[00:50:59]For B C D I will take moments about B.

[00:51:03]When you take moments about B for B C D T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T * the perpendicular distance to T 2 A sin alpha 2 A sin alpha and again minus lambda W * A sin alpha - 2 lambda W - 2 lambda W time the perpendicular distance to 2 lambda W that 2 A sin alpha A sin alpha that comes Tree a sin alpha tree a sin alpha equals par. Now here you know that when you take moments about b lambda 2 lambda w and t that's it that because p passes through pal a sin alpha to a sin alpha to a sin alpha cancel you will get that t equal t= here 6 + 7 7 lambda w / 2 t = 7 lambda w / 2 Right now the necessary step is what?

[00:52:13]Now we have what is t? The necessary step is what? You can take moments about a for a b c d or right for a b comma b c comma c d about a moment.

[00:52:33]Now when you take moments about a for a b c d right these forces are not coming that because these are internal forces.

[00:52:43]Now t rotate in this way and even p rotates in this way t time what was t 7 lambda w / 2 time perpendicular distance to t from a 2 a sin alpha 2 a sin alpha that comes 4 a sin alpha 4 a sin alpha plus p was slammed W tan alpha lambda w tan alpha lambda w tan alpha time lambda w tan alpha time what's the perpendicular distance to p perpendicular distance to p is 2 a cos alpha 2 a cos alpha minus now from this one this one and this one the rotation is in the other way around minus minus w * a cos alpha Uh nous w * a sin alpha alpha a sin alpha minus w * a sin alpha minus lambda w * 3 a sin alpha minus 2 lambda w 2 lambda w * the perpendicular distance to 2 lambda w here it comes 2 a 2 a A 5 A sin alpha 5 a sin alpha equal zero. Now you can see from all the terms that a sin alpha 2 actually w a sin alpha to w a sin alpha to w a sin alpha will be cancel tan alph sign over cos then cos 2 cos cancel right then sign is remaining cancel finally we will get that see 7 into 2 right that's come 14 14 lambda 14 lambda now here I see 2 lambda plus 2 lambda See then min - w or here - w - w now here - 3 lambda - 3 lambda and here - 10 lambda no - 10 lambda equ= 0 what will happen now here -3 now here that 16 - 13 that okay uh this this should be one right here one lambda= half equal half. Oh, one wait 14 and 16 16 - 13 1 / 3 right lambda= 1 / 3 lambda= 1 lambda= 1 lambda= 1 that you can find what is t and p right lambda= 1 what tal 7 w 7 lambda w /2 lambda is 1 / 3 7 w / 6 7 W / 6 and the other thing now this is the updated value and the other thing what is P that P we have taken lambda W tan alpha lambda is 1 / 3 then W tan alpha / 3 update since we know what is T and P you can find reaction P the horizontal and vertical component of the reaction P then now when You consider that B C and C D you can see that the only okay horizontal forces are what P and this one P like this what can we think about the Y component lambda W 2 lambda W this is 3 lambda W 3 lambda W lambda was 1 / G and W downward w downward you know that's T was 7 W 6 7 W right.

[00:57:01]Okay.

[00:57:04]Now what can you do is you can resolve forces for BC CD two rods horizontally and vertically right now YB downward therefore YB upward like this see for BC comma CD for BC comma CD we can resolve forces in this manner. Write W tan alpha / 3 - XB= 0. XB equal W tan alpha / 3 my horizontal component at reaction at B horizontal component at capital B and again uh for BC comma CD for BCA CD you can resolve forces upward T 7 W / 6 minus lambda W / lambda w / 3 - 2 w / 3 - yb=0 here you will get that yb= w / 6 okay this is the vertical component at b vertical component at capital B Okay. Right.

[00:59:02]We got the answer.