Statically Indeterminate Beams

Transcription

1 Deflection Part Staticall Indeterminate eams We can use the same method that we used for deflection to analze staticall indeterminate beams lessed are the who can laugh at themselves for the shall never cease to be amused. Wednesda, November, Meeting Thirt Five Staticall Indeterminate eams If we start with a beam loaded as shown The left end is supported as a fixed end support The right end is supported on a roller Staticall Indeterminate eams If we remove the supports and look at the reactions we have M x Wednesda, November, Meeting Thirt Five Wednesda, November, Meeting Thirt Five 4 1

2 Staticall Indeterminate eams We have four unknowns and three equilibrium conditions Staticall Indeterminate eams We use up one of those conditions solving for x = M M x Wednesda, November, Meeting Thirt Five 5 Wednesda, November, Meeting Thirt Five 6 Staticall Indeterminate eams We still have three unknown reactions and onl two equilibrium conditions to solve for them Staticall Indeterminate eams We can reduce the distributed load to a point load but that probabl won t help much M M Wednesda, November, Meeting Thirt Five 7 Wednesda, November, Meeting Thirt Five 8

3 Staticall Indeterminate eams We do know the loading rate on the beam so we can start from that M = w w x Staticall Indeterminate eams From that we can look at the change in the shear across the beam M w x = w dv = w( x) = w dx Wednesda, November, Meeting Thirt Five 9 Wednesda, November, Meeting Thirt Five 1 Staticall Indeterminate eams Integrating to find the shear at an distance into the beam we have V( x) = + C1 Staticall Indeterminate eams We have two possible boundar conditions that we can use to solve for C 1 = + 1 V x C at x =, V = and x = V, = M M Wednesda, November, Meeting Thirt Five 11 Wednesda, November, Meeting Thirt Five 1

4 Staticall Indeterminate eams Using the left hand boundar condition we have M 1 V x = + C 1 V() = = + C C = 1 Staticall Indeterminate eams So our expression for the shear is M V ( x) = + Wednesda, November, Meeting Thirt Five 1 Wednesda, November, Meeting Thirt Five 14 Staticall Indeterminate eams If we now utilize the expression for the shear to develop the expression for the moment at an x we have V x = + dm = V ( x ) = + dx = + x + C M Wednesda, November, Meeting Thirt Five 15 Staticall Indeterminate eams We can use our boundar condition again to solve for C M = + x + C at x =, M = M Notice that the sign of M is negative. This is due to the w fact that we are M = M = + + C directing our -axis M = C Wednesda, November, Meeting Thirt Five 16 downward so a positive moment will be into the page or CW. (i cross j) = positive k 4

5 Staticall Indeterminate eams So our expression for the moment at an point in the beam is M Staticall Indeterminate eams We can continue with the process utilizing our slope equation = + x M = + x M dθ = dx EI M Wednesda, November, Meeting Thirt Five 17 Wednesda, November, Meeting Thirt Five 18 Staticall Indeterminate eams n easier to use form can be generated b multipling both sides b EI and taking the negative sign inside of the expression for the moment Staticall Indeterminate eams Integrating for the slope we have M M dθ EI =+ x + M dx θ =+ 6 x EI x + M x + C Wednesda, November, Meeting Thirt Five 19 Wednesda, November, Meeting Thirt Five 5

6 Staticall Indeterminate eams We have a boundar condition that we can utilize here to solve for C t the support at, the slope is equal to Staticall Indeterminate eams So the slope equation has the form x =+ + + EIθ x M x C 6 w EIθ = =+ + M + C 6 = C Wednesda, November, Meeting Thirt Five 1 M x EIθ ( x) =+ + M x 6 Wednesda, November, Meeting Thirt Five M Staticall Indeterminate eams Finall utilizing the deflection relationship we have Staticall Indeterminate eams Utilizing our boundar condition at the support at to solve for C 4 t the support we have a deflection of M M dv θ ( x) = dx 4 x M x EIv( x) =+ + + C x M x EIv( x) =+ + + C ( ) M 4 w EIv = =+ + + C 4 6 = C 4 Wednesda, November, Meeting Thirt Five Wednesda, November, Meeting Thirt Five 4 6

7 Staticall Indeterminate eams So our final form for the deflection is EIv x x 4 M x =+ + Wednesda, November, Meeting Thirt Five 5 M 4 6 Staticall Indeterminate eams There are actuall still two boundar conditions that we know but we haven t used t x =, M= t x =, v = x EIv( x) = = + x M Wednesda, November, Meeting Thirt Five 6 M 4 M x Staticall Indeterminate eams If we use these two conditions in the expression below, we will have two equations with two unknowns t x =, M= t x =, v = EIv Wednesda, November, Meeting Thirt Five 7 M 4 w M = = w = = + M M Staticall Indeterminate eams Working on the math w 1 =+ + w = + M Wednesda, November, Meeting Thirt Five 8 M M 7

8 Staticall Indeterminate eams dding the equations together 5w = + 1 5w 5w = = 1 8 Wednesda, November, Meeting Thirt Five 9 M Staticall Indeterminate eams M Substituting the into the second equation we have = + w 5w ( ) M 8 = + w M w 5w w = + = 8 8 Wednesda, November, Meeting Thirt Five M Staticall Indeterminate eams You can substitute the values for M and in to the expressions for slope and deflection You can also solve for using the equilibrium expression w 5w w M = + = 8 8 Wednesda, November, Meeting Thirt Five 1 M oundar Conditions If ou can write the expressions for the loading, shear, moment, slope, and deflection ou have ver powerful tools for solving indeterminate structures To be able to use these tools ou must be able to identif the boundar conditions that can be used Wednesda, November, Meeting Thirt Five M 8

9 oundar Conditions oundar Conditions If ou are at a fixed end support The slope (θ) is equal to If ou are at a point in a beam where two solutions meet The deflection (v) is equal to If ou are at a pin or roller The deflection (v) is equal to The moment (M) is equal to M The slope (θ) from one direction is equal to the slope from the other direction The deflection (v) from one direction is equal to the deflection from the other direction The moment (M) from one direction is equal to the moment from the other direction The shear (V) from one direction is equal to the shear from the other direction Wednesda, November, Meeting Thirt Five Wednesda, November, Meeting Thirt Five 4 Homework In Class test next Wed Wednesda, November, Meeting Thirt Five 5 9