Torsion Part 3 n honest man can feel no pleasure in the exercise of power over his fellow citizens. -Thomas Jefferson In a manner similar to that we used when dealing with axial loads and statically indeterminate structures, we will use our relationships between applied moments and the angle of twist to solve for reactions that would be statically indeterminate 3 September 22 Statically Indeterminate Torsion 2 In we look at a simple system of a bar that has a fixed end support at each end and a moment applied somewhere along the length of the beam We know that the fixed end supports provide resistive moments as well as reactive forces Since there are no applied forces to react to (assume the weight of the bar is negligible) there are no force reactions at the end 3 September 22 Statically Indeterminate Torsion 3 3 September 22 Statically Indeterminate Torsion 1
If we draw a FD of the bar system removing the supports at the end we have T and T are the assumed support moments/torques at the ends of the bar If we use our equation of equilibrium can write the expression M M T + T T 3 September 22 Statically Indeterminate Torsion 5 3 September 22 Statically Indeterminate Torsion 6 This expression alone is not sufficient to solve for the reactive torques, T and T M M T + T T We will need to use the material response characteristics to determine these support moments M M T + T T 3 September 22 Statically Indeterminate Torsion 7 3 September 22 Statically Indeterminate Torsion 8 2
If we look at the front section of the beam, the section labeled, we can pass a plane through the bar just before the point where the applied moment, T, is applied If we look at the FD of that section alone, we will need a torque at the end away from the wall that is equal and opposite T M M T + T T M M T + T T 3 September 22 Statically Indeterminate Torsion 9 3 September 22 Statically Indeterminate Torsion 1 If we label that torque at T then the magnitude of T would be equal to T but the sense of rotation would be the opposite If the section we cut through the bar is just before we get to the applied torque, T, then the distance that this torque is from the wall is equal to L M M T + T T M M T + T T 3 September 22 Statically Indeterminate Torsion 11 3 September 22 Statically Indeterminate Torsion 12 3
The angle of twist a the plane in reference to the wall can then be calculated as 3 September 22 Statically Indeterminate Torsion 13 Now if we move our plane to just the other side of the applied moment, we will have a resistive moment that has a magnitude of T which is equal to T T T 3 September 22 Statically Indeterminate Torsion 1 Since these two planes are almost coincidental, the angle of twist will be the same for any point along common radii on these planes T T T ( ) T T L 3 September 22 Statically Indeterminate Torsion 15 There we can set the two expressions for the angle of twist equal to one another (they are viewed from different reference planes) ( ) T T L 3 September 22 Statically Indeterminate Torsion 16
Doing the algebra G J L L T T L L 3 September 22 Statically Indeterminate Torsion 17 nd simplifying we have T L L L T T L L You are quite welcome to take it even farther if you wish. 3 September 22 Statically Indeterminate Torsion 18 Given: Two bars, and as shown ar has a diameter of 1 in, a shear modulus of 6x1 6 psi, and a length of 1 in ar has a diameter of 2 in, a shear modulus of 3.6x1 6 psi, and a length of 1 in The bars are originally separated by a small (will not affect the length of the bars) gap 3 September 22 Statically Indeterminate Torsion 19 L : 1in L : 1in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in The only reason I am using for pounds force is that MathCad differentiates between pounds force and pounds mass. Given: torque is applied to bar rotating it about its axis 2 and then the bars are welded together Required: What are the magnitude of the angles of twist of the two bars afterward? 2 o o 3 September 22 Statically Indeterminate Torsion 2 L : 1in L : 1in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in 5
Solution: Once the bars are welded together, the angle of twist for each bar will be the same at the point where they are joined. We will given this final angle of twist the symbol L : 1in L : 1in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in Solution: The torque that was initially applied to the bar can be found by using the formula for predicting the twist from the material properties L : 1in L : 1in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in 3 September 22 Statically Indeterminate Torsion 21 3 September 22 Statically Indeterminate Torsion 22 Solution: We know everything in this formula except the polar moment of inertia and since the cross section of the rod is solid, we can easily calculate that J π d 2 2 L : 1in L : 1in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in π d J : 2 2 J.98 in Solution: Since we know all the material properties, we can calculate the torque which was required to produce the original angle of twist T T 2.56x1 3 in-lb L : 1in L : 1 in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in π d J : J.98 in : 2deg G J T 2.56 1 3 in L 3 September 22 Statically Indeterminate Torsion 23 3 September 22 Statically Indeterminate Torsion 2 6
Solution: We can say that the torque at the wall support is T and that the torque at the wall support at is T We can also assume that both torques are opposite in sense to the applied torque For equilibrium M T T + T 3 September 22 Statically Indeterminate Torsion 25 T T 2.56x1 3 in-lb L : 1in L : 1 in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in π d J : J.98 in : 2deg G J T 2.56 1 3 in L Solution: If we now calculate the angle of twist in the section at the point where the bars join assigning the contact point at the wall as the reference we have Now that the bars are joined the angle of twist is no longer 2 degrees 3 September 22 Statically Indeterminate Torsion 26 T T 2.56x1 3 in-lb L : 1in L : 1 in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in π d J : J.98 in : 2deg G J T 2.56 1 3 in L Solution: Working on the section again assigning the end at the wall as the reference we can say G J T T 2.56x1 3 in-lb L : 1in L : 1 in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in π d J : J.98 in : 2deg G J T 2.56 1 3 in L Solution: Since the bars are joined together, the angles of twist have to be equal to and opposite in sense from each other (consider the reference points) T T 2.56x1 3 in-lb L : 1in L : 1 in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in π d J : J.98 in : 2deg G J T 2.56 1 3 in L 3 September 22 Statically Indeterminate Torsion 27 3 September 22 Statically Indeterminate Torsion 28 7
Solution: Utilizing our equilibrium condition developed T earlier we have 2.56x1 3 in-lb T T T ( ) T T L L : 1in L : 1 in G 6 1 6 : G 3.8 1 6 d : 1in d : 2in π d J : J.98 in : 2deg 3 September 22 Statically Indeterminate Torsion 29 T G J T 2.56 1 3 in L Solution: Solving for T 2.56x1 3 in-lb + L L + T L G : 6 1 6 d J : : 1in 1in π d J.98 in : 2deg ( G J) L ( T L) TTOP : ( G J) 3 September 22 Statically Indeterminate Torsion 3 L d J : : : 2in T 1in G : 3.8 1 6 π d J 1.571 in T 2.56 1 3 in TTOP 3.5 1 3 L L TOTTOM : + TOTTOM 1.865 1 5 ( in) 1 G J G J TTOP T 18.69in TOTTOM T T T 1.871 1 3 in T Solution: Then solving for T 2.56x1 3 in-lb J.98in : 2deg G J L T L T TOP : ( G J ) L L T OTTOM 1.865 1 5 1 T OTTOM : + ( in ) G J G J T TOP T : T 18.69in T OTTOM T : T T T 1871.8in J 1.571in T 2.56 1 3 in T TOP 3.5 1 3 3 September 22 T L Statically Indeterminate Torsion 31 T T Solution: Using our T to solve for 3 September 22 Statically Indeterminate Torsion 32 T T 2.56x1 3 in-lb L L T OTTOM : + T OTTOM 1.865 1 5 1 ( in ) G J G J T TOP T : T 18.69in T OTTOM T : T T T 1871.8in T L :.18deg G J T L :.18deg G J 8
Solution: Checking by solving for T T 2.56x1 3 in-lb L L T OTTOM : + T OTTOM 1.865 1 5 1 ( in ) G J G J T TOP T : T 18.69in T OTTOM T : T T T 1871.8in T L :.18deg G J T L :.18deg G J Solution: The only problem we have is that section started from a 2 angle and what we solved for was actually the change in angle when was attached to The actual twist angle (from the unloaded condition) for would be the 2 angle minus this new angle of twist T T 2.56x1 3 in-lb : T L.18 deg G J T L :.18deg G J : 2deg 1.82 deg 3 September 22 Statically Indeterminate Torsion 33 3 September 22 Statically Indeterminate Torsion 3 Your turn PROLEM -3.11 Homework No homework due Monday, just the take home portion of the class due at the beginning of the class period In class, multiple choice part of the test on Monday 3 September 22 Statically Indeterminate Torsion 35 3 September 22 Statically Indeterminate Torsion 36 9