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calc. reactions for propped cantilever
- Thread starter nyvens
- Start date
When you say a propped cantilever, I am picturing a single span beam that is fixed at one end and pinned at the other. The steel manual has this as case 12 in the 13th edition manual (pg 3-214).
The reaction at the pin support is 0.375wl, the reaction at the fixed support is 0.625wl.
The reaction at the pin support is 0.375wl, the reaction at the fixed support is 0.625wl.
Assume the support is fixed. And the cantilever subjects to uniform distributed load (UDL) "w".
The reactions at a distance "x" from the free end are simply:
Shear (V) = w*x
Moment (-M) = -w*x*(x/2) = -w*(x^2)/2
(-) means tension on the loading side of the cantilever.
Without external loads, the cantilever will have reactions due to self-weight alone.
Hope I understood your question, and have answered correctly.
The reactions at a distance "x" from the free end are simply:
Shear (V) = w*x
Moment (-M) = -w*x*(x/2) = -w*(x^2)/2
(-) means tension on the loading side of the cantilever.
Without external loads, the cantilever will have reactions due to self-weight alone.
Hope I understood your question, and have answered correctly.
- Thread starter
- #5
without getting into Castigliano's method (my favorite) or some other indeterminate analysis, this is what I would do (if I had to do it by hand).
1. Remove the support at S, and calc the deflection of the beam at that location. See 13th edition manual case 19.
Also calc the moment at the support under this condition.
2. Use case 21 in the 13th ed manual and apply a point load P at the location of the reaction S (using only that point load and not the UDL). Determine what that force (reaction) needs to be to make the upward deflection equal to the downward deflection of the UDL.
3. Calc the moment of the reaction S (and not the UDL) at the fixed support.
4. Your reaction S is already calc'ed (see step 2). The vertical reaction at R is done by summing forces in the y direction. The "built-in moment" (I've never heard that terminology used before), is the algebraic sum of the UDL only moment and the reaction,S, only moment.
Hope that helps.
1. Remove the support at S, and calc the deflection of the beam at that location. See 13th edition manual case 19.
Also calc the moment at the support under this condition.
2. Use case 21 in the 13th ed manual and apply a point load P at the location of the reaction S (using only that point load and not the UDL). Determine what that force (reaction) needs to be to make the upward deflection equal to the downward deflection of the UDL.
3. Calc the moment of the reaction S (and not the UDL) at the fixed support.
4. Your reaction S is already calc'ed (see step 2). The vertical reaction at R is done by summing forces in the y direction. The "built-in moment" (I've never heard that terminology used before), is the algebraic sum of the UDL only moment and the reaction,S, only moment.
Hope that helps.
a proped cantilever (as you've drawn it) is a singly redundant beam. look up "unit load method". or look up "Roark", this is a standard problem.
looking at the other respondants, kslee1000 is given you pointers for a determinate beam which won't work out. StructuralEIT's first post may be right for the shear reactions but lacks the moment (which you could calculate, if the shear's are right, but i suspect they are for a cantilever propped at the end).
looking at the other respondants, kslee1000 is given you pointers for a determinate beam which won't work out. StructuralEIT's first post may be right for the shear reactions but lacks the moment (which you could calculate, if the shear's are right, but i suspect they are for a cantilever propped at the end).
a picture/sketch better than thousand words!
Lets lable the fixed end as "B", the simple support as "A", the distance from free end to the simple support as "a", the distance in between supports as "b", the total beam length as "L = a+b". Now solving the reaction at the simple support by consistent displacement method:
1. Displacement at "A" for a cantilever with legth "L" and UDL "w" equals:
w*(a^4-4aL^3+3L^4)/(24EI)
2. Displacement at "A" for the same cantilever with a concetrate load "P" equals:
P*b^3/(3EI)
3. Equate displacements from 1 & 2 you can solve force P, which is the reaction of support A. From there on, you can solve the reactions at the fixed support by simple statics.
Above equations can be found at many structural analysis text books, as well as the AISC beam manual. Please verify.
Lets lable the fixed end as "B", the simple support as "A", the distance from free end to the simple support as "a", the distance in between supports as "b", the total beam length as "L = a+b". Now solving the reaction at the simple support by consistent displacement method:
1. Displacement at "A" for a cantilever with legth "L" and UDL "w" equals:
w*(a^4-4aL^3+3L^4)/(24EI)
2. Displacement at "A" for the same cantilever with a concetrate load "P" equals:
P*b^3/(3EI)
3. Equate displacements from 1 & 2 you can solve force P, which is the reaction of support A. From there on, you can solve the reactions at the fixed support by simple statics.
Above equations can be found at many structural analysis text books, as well as the AISC beam manual. Please verify.
MiketheEngineer
Structural
Run it through RISA or other software
There are many ways of doing it. Let's say the span is L and the cantilever is C.
The fixed moment of a beam without a cantilever is wL^2/8.
The cantilever moment is wC^2/2.
Using superposition, the support moment must be wL^2/8 - wC^2/4.
From that, the reactions can be readily determined.
Best regards,
BA
The fixed moment of a beam without a cantilever is wL^2/8.
The cantilever moment is wC^2/2.
Using superposition, the support moment must be wL^2/8 - wC^2/4.
From that, the reactions can be readily determined.
Best regards,
BA
so much discussion on a prop cantilever.
aaannnnnddddd.....here's another one:
Mmax (@ fix support) = wl^2/8
Mmax (@ Like 0.4L away from fix support) = 9/128wl^2
Vmax (@ fix support) = .625wl
someone correct me if i'm wrong, but these are the equations i remember off the top of my head without having the steel code in front of me. unless we're discussing something else.
or i'd do it with moment distribution. which i probably need to brush up on anyway.
aaannnnnddddd.....here's another one:
Mmax (@ fix support) = wl^2/8
Mmax (@ Like 0.4L away from fix support) = 9/128wl^2
Vmax (@ fix support) = .625wl
someone correct me if i'm wrong, but these are the equations i remember off the top of my head without having the steel code in front of me. unless we're discussing something else.
or i'd do it with moment distribution. which i probably need to brush up on anyway.
swivel63,
Except for the location of the maximum positive moment, your memory is pretty good...for a beam fixed at one end and hinged at the other.
But we are considering a beam with a fixed end, a roller support and a cantilever beyond the roller support.
As you have noted, the fixed end moment for the uniform load in the span portion of the beam is wl2/8. The cantilever moment at the roller support is wc2/2 which produces a moment of -wc2/4 at the fixed support (where c is the length of cantilever). Combining these results produces a fixed end moment of wl2/8 - wc2/4.
The reaction at the roller support can be found from:
Rroller = w(l+c)2/(2*l) - Mfixed/l which agrees with the method proposed by GerhardSA.
I prefer this method because it is easily remembered and does not rely on accessing a formula which may not be immediately at hand.
Best regards,
BA
Except for the location of the maximum positive moment, your memory is pretty good...for a beam fixed at one end and hinged at the other.
But we are considering a beam with a fixed end, a roller support and a cantilever beyond the roller support.
As you have noted, the fixed end moment for the uniform load in the span portion of the beam is wl2/8. The cantilever moment at the roller support is wc2/2 which produces a moment of -wc2/4 at the fixed support (where c is the length of cantilever). Combining these results produces a fixed end moment of wl2/8 - wc2/4.
The reaction at the roller support can be found from:
Rroller = w(l+c)2/(2*l) - Mfixed/l which agrees with the method proposed by GerhardSA.
I prefer this method because it is easily remembered and does not rely on accessing a formula which may not be immediately at hand.
Best regards,
BA
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