BOLTED CONNECTION 2

追加: 2026-05-07

[00:00:01]Hello everyone, welcome to another video in our civil engineering review and refresher.

[00:00:09]So for this video, this features eccentrically loaded bolted connection in steel design.

[00:00:19]So let us say we have a group of bolts and then this is represented by three bolts. Here we have bolt number one, bolt number two and bolt number three with their center of gravity as this point. So the horizontal axis we can take this as the x-axis and the vertical axis passing through the centrid we can call this as the y-axis.

[00:00:49]And then this group of bolts is loaded by an eentric load P at an angle of theta with the horizontal as shown in the figure.

[00:01:01]So this distance is what we call the eentricity with respect to yaxis. So we have e x and this is the eentric distance with respect to x. So we have ey.

[00:01:17]Then what are the effects of the load to the group of bolts? So number one it affects the bolts by direct shear and number two since the load P is not acting at the center of gravity it will produce torsion. So we have also the torsional shear as the second effect.

[00:01:44]So we will take up each separately.

[00:01:48]So a if we have the direct shear.

[00:01:52]So for direct shear let us assume that the load P is being distributed equally to all the bolts. So for our load P we can transfer it to the center of gravity and then we can resolve the P in terms of the horizontal component PX which is equal to P's cossine theta in this figure and a vertical component of P we have Py equal to P sin of theta and then the bolts will react also since the bolts are acted by PX and PY. We have also the reaction PDX and PD Y and again the PDX and PDY to all the bolts are equal because it is assumed that the load P is equally distributed to all the bolts. So from the figure we can have pdx is simply equal to px divided by the number of bolts n and pdy is equal to py divided by the number of bolts n. So we have there and then we consider letter b the torsion effect of the load to the bolts.

[00:03:21]So we have P at an angle of theta and this is at eentric distances E X and E Y as shown in the figure. So we can resolve the P in terms of the components. We have the px and the py.

[00:03:38]And the px and the py with distances eentricity with respect to y or ey and e respectively will produce a twisting moment or torsion about the center of gravity. And that torsion t is equal to px * ey plus py * e x.

[00:04:03]uh then what are the formulas?

[00:04:07]So uh the bolts will react to the tone ation. So we have ptx and then pt y. Now take note that the ptx and pty in each bolts are not equal because we know in rotation or in twisting the farthest element from the center of gravity will experience more stress compared to the elements nearer from the center of gravity.

[00:04:44]So if we consider the three bolts here, bolt one, bolt two, bolt three, we can say that bolt one is the most stressed bolt compared to bolt two and bolt three.

[00:05:00]So how will we solve for the ptx and pty?

[00:05:05]So if if we consider bolt one which is the most stressed in torsion we have the distance of the bolt one from the x and y axis respectively. So we have distance y vertically and distance x horizontally.

[00:05:23]So ptx is equal to t * y / j. So t * y / j and pt y is equal to t * x / j.

[00:05:39]And what is j? Uh we know if we recall we know t is torsion. Y is the distance of the bolt and also x the distancees of the bolt from the coordinate axis. And then J is what we call the polar moment of inertia.

[00:05:58]So without derivation we can compute the polar moment of inertia of the bolt of group of group of bolts. I mean polar moment of inertia of the group of bolt by using the formula summation of x^2 + y^ 2.

[00:06:18]So this is summation of x^2 + y^ 2.

[00:06:24]And then again we will note that the farthest bolt from the center of gravity is the most stressed bolt in torsion.

[00:06:35]Okay. Then we consider the most stressed bolt that is bolt number one. So it has pdx it has pdy by direct shear. It has also ptx and pty by torsion. And then we can have the resultant force r act acting on bolt number one or the reaction of bolt number one. And that is r which is obtained by pythagorian theorem. R is square root of PTX + P DX quantity squared plus PTY + PDY quantity squared.

[00:07:20]Okay. So we can now apply our formulas by solving problems.

[00:07:28]So situation one figure ST 231 shows an eccentrically bolted connection.

[00:07:38]The applied load is P equal to 60 konton. The bolt diameter is 19 mm.

[00:07:47]Question number one. Determine the shear force of the most critical bolt based on direct shear alone.

[00:07:56]So direct shear that means no torsion considered. Then number two determine the sheer force of the most critical bolt based on torsion alone. So number two based on torsion alone meaning no direct shear considered. And then number three determine the maximum shear stress of the most critical bolt due to the given loading. So number three is the combination now of the two. So we have the direct shear in number one and the torsion stress in number two. Uh so we have this we have six bolts in the connection then we the given distances and then loaded by a vertical load P and that is equal to 60 kilo newtons.

[00:08:51]Okay. So we will now solve by using direct shear or solve number one and that is direct shear. So solution as if we transfer the 60 konton to the center of gravity. Uh by the way our bolts here the group of bolts they are symmetrically arranged. So obviously the center of gravity falls at the geometric center. So we have here the center of gravity of the bolt group.

[00:09:28]Okay. So notice that the load is purely vertical. So that means the load P which is 60 automatically this is equal to P Y and PX is equal to zero because there is no horizontal component of the 60 kon newton load.

[00:09:50]So for direct share there is no pdx only pdy.

[00:09:57]So if we solve pdy equal to the py which is equal to p / n. So 60,000 / 6. So PDY is equal to 10,000 Newton.

[00:10:10]So the stress is PDY over the area. So PDY is 10,000. Our bolt is 19 mm. So pi over 419².

[00:10:24]So the shear of the bolts due to direct shear is 35.27 27 megapascals.

[00:10:37]Question number two, what is the force of the bolts due to torsion? Uh the force in the most critical bolt due to torsion.

[00:10:48]So notice that in the arrangement of bolts, bolt number one, bolt or bolt number one, this bolt, let us say bolt number three, bolt number four and bolt number six are the farthest from the center of gravity.

[00:11:08]So I will just consider bolt number one as the most stressed bolt.

[00:11:15]So bolt number one has ptx pty and the resultant force which is the question the pt. So again the force on the most critical bolt on torsion.

[00:11:32]Okay. So we have p equal to 60 konton and this is purely vertical. So py is 60 while px is zero.

[00:11:45]So our torsion which will be developed at the center of gravity equal to px ey plus py e x. So there is no px. So simply we have py 60 * e x which is equal to 200.

[00:12:05]Uh by the way this is how come we get 200 from the figure. So balika figure.

[00:12:16]Okay. So half is 60 + 40. So we have 100 plus another 100 here. So we have 200.

[00:12:28]Okay. So 60 *2 that is 12 konme.

[00:12:33]The polar moment of inertia of the group bolts or the bolt group we have summation of x^2 + y^2 where x and y are the distances of each bolt from the center of gravity. Okay. So for this group of bolts considering this as a center of gravity now. So this is at 60 and then we have 100. So if I will have the X dens distances they have equal X distances from the center of gravity and that is 60.

[00:13:08]Okay. So we have also 60 here. 60 60 also on the other side. So the summation of X squ so we have 6 times the distance 60 squared. And then for the distances of y we have 1 2 3 4 with y = 100. So 4 * 100².

[00:13:31]How about these two bolts here? So these two bolts have no y distances because they lie on the center of gravity or they lie on the horizontal axis with the center of gravity. So the summation of x^2 + y^2 or that is the j equal to 61,600.

[00:13:56]So I can solve for ptx and pty from the formula.

[00:14:01]So pty and then ptx.

[00:14:05]So ptx is ty / j where we use the x and y for our bolt number one. Uh so for bolt number one x is 60 same as this and then y is 100. I just place x and y here so that it will not complicate the drawing in the figure. So t is 12,000 then y is 100 over 61,600 19,480.

[00:14:38]Okay. Then pty we have t * x over j. So t is 12,000, tx is 60 and then j is 61600.

[00:14:57]So if you compare ptx and pty the same l t / j they only differ in y and x. So ptxty / j pty tx / j. So 11688.

[00:15:15]So after that we can solve for the force pt by pythagorian theorem. So pt equal to the square root of ptx² + pty squared. So with ptx and pty values we have 22.72.

[00:15:37]Problem number three, what is the stress in the most critical bolt?

[00:15:44]So again by the combination of our direct shear and torsion, the most critical bolt is bolt number one. And for bolt number one, we have ptx. We do not have pdx.

[00:16:01]We have pdy and then we have pty and then we have the r the total resultant.

[00:16:08]So r is ptx + pdx² where pdx is zero plus pty + pdy squared and then extract the square root. So r is 29152.

[00:16:22]So the total stress of the bolt by direct shear and torsion is now equal to r / a. So 29152 over the area of one bolt. The bolt number one. So 102.82 mega pascals.

[00:16:49]Situation number two.

[00:16:51]The gaset plate connection shows six 16 mm diameter bolts supporting the load P as shown in the figure.

[00:17:03]So the P is 82 konton.

[00:17:06]S1 equal to 50, theta is 80, S_UB_2 is 75, S3 is 90, and S4 is 250.

[00:17:21]So number one determine the shear stress develop in the bolts due to direct shear only. Number two determine the maximum shear stress develop in the most critical bolt due to torsion only. And number three the allowable bolt shear stress is 115 megapascal. Determine the minimum required bolt diameter to prevent overstressing.

[00:17:47]So as you notice our questions one, two and three the same lang in our first situation problem solved. So number one is due to direct shear only number two is due to torsion only and number three is due to the given loading which is the combination of the two.

[00:18:13]Now notice that our bolts here are not symmetrically laid. So we need to look for the center of gravity. And the center of gravity can be solved by using our area moment theorem.

[00:18:30]So if we let a equal to the area of one bolt and we can have this as our reference.

[00:18:39]So the edges now so from the left edge and then the upper top of the bolt group. So I have here the distance bar x for the center of gravity and the distance bar y for the center of gravity.

[00:18:58]So by area moment the total area is 6 * a and then multiplied by bar x equal to the summation of the x distances of the bolts from the reference.

[00:19:13]So we have here area total bar x equal to the summation of area time x. So 6 a bar x. Now notice from this reference we have dawang or we have two bolts with distances 90. So that is why we have 2 * 90. By the way this three bolts has zero distances from the reference vertical axis. So 2 * 90 + 1 bolt. The distance here is 9090 or 180. So bar x is 60.

[00:19:48]For bar y we have this vertical uh horizontal line as reference.

[00:19:56]So area bar y equal to summation of area y.

[00:20:01]So 6 a bar y. So as reference we have two bolts with distance 75 and then we have one bolt with distance 150. Okay.

[00:20:12]Uh I will correct this because I have uh this is bar y. Sorry for the misprint misprint talaga.

[00:20:39]Okay. So that is bar y equal to 50.

[00:20:44]Okay. So for number one this is direct shear. So as if we transfer our load p here and then have the components px and p y. So p is 80 if you recall and the angle theta is 80 with respect to the horizontal.

[00:21:05]So it reacted. So the bolts reacted by pdx and pdy equally of all the bolts. So px is p cosine theta. So 82 cosine 80. So 1424 and py is p sin theta. So we have 80.75.

[00:21:28]Okay. So pdx is px / n because px is equally divided. So 1424 / 6 that is 2.373.

[00:21:40]pdy is also pyide by n. So 8075 / 6. So 13.458.

[00:21:49]So if I will consider one bolt because they have equal pdx and pdy. I will just consider this one bolt. Here we have the PD the total force. So PDX plus PDY. So we have the values PD is 1367.

[00:22:07]So we have 16 mm bolt. So I can solve the shear stress by using the force PD divide by the area. So PD is 13670 / 416².

[00:22:21]So 68 mega pascals.

[00:22:27]So number two, so sheer stress of the most stressed bolt or more most critical bolt due to torsion only.

[00:22:37]Okay. So we will not transfer the force to the center of the gravity but we will have the torque develop at the center of gravity. So px and py if we recall the values we have this. And then notice we have 250. If you recall this is 250 and this bar x is 60. So we have here ex equal to 190. By the way ex is from center of gravity.

[00:23:11]Then if you recall this is 50 here and bar y is also 50. So we have ey equal to 100. So the torque developed at the center of gravity is px ey + py x. So 14.24 *.1 1875 *.19.

[00:23:34]So 610 7665 kon newton meter. Then we have the J.

[00:23:41]So notice that if you have the X we have three bolts with X equal to 60 because bar X is 60.

[00:23:51]This is 90. So that is why we have here 3 * 60 squared.

[00:23:55]The uh distance here is 90 and X is 60.

[00:23:58]So we have here 30. So we have two bolts with X distance is 30² and this is 90 + 30. So we have one bolt with distance 120.

[00:24:13]Then we have the Y distances of the bolts from the CG. Bar Y is 50. So we have three bolts with Y distance 50.

[00:24:24]Then this is 75. Bar Y is 50. So we have here 25. So we have two bolts with Y distances 25².

[00:24:34]Then this is 75 and then 25. So we have one volt with Y distance 100.

[00:24:44]So J is 45750.

[00:24:49]Uh so we can now solve for the PT. Uh by the way from this our most stress bolt is bolt number one farthest from the center of gravity.

[00:25:03]Okay. So we can now solve for the ptx and pty of bolt number one. So by the way this distance bolt number one from cg this is 120 90 + 30 and this is 50 considering bolt number one. So we have this as bolt number one.

[00:25:32]The distance x of bolt number one from the center of gravity is 120 and the distance y of bolt number one from the center of gravity is equal to 50 mm.

[00:25:46]So we have now the ptx and the pty. So if we solve for ptx that is t * y / j and substituting values for t y and j.

[00:26:00]So we have 18 324 same as pty. So that is equal to tx / g.

[00:26:08]So substituting the values with x = 120.

[00:26:13]So we have pty 43,978 ntons.

[00:26:18]Then we have the resultant PT.

[00:26:22]So the resultant force of bolt number one due to torsion. So square root of ptx² + pty squared with the values computed we have pt equal to 47 642 newtons then the stress of the most stressed bolt which is one bolt number one due to torsion is pt / a so 47642 over the area of 16 mm so that is 236.96 megapascals Question number three. So if the allowable bolt shear stress is 115 megapascal, determine the minimum required bolt diameter to prevent overstressing.

[00:27:12]So bolt one is the most stressed bolt.

[00:27:17]So we have the ptx, pdx, pty and then pdy. Then we have the resultant.

[00:27:25]So for R it is the combine the combination of the two effects. Then substituting values we have R equ= to 61051.

[00:27:37]Then for the stress that is R / A. So with the allowable stress 115 megapascal we can solve for D equal to 26 mm.

[00:27:51]Okay. Thank you very much for viewing the video and I hope my video helped you in your studies. Stay tuned for the next videos to come.