7-49 Draw the Shear and Moment Diagrams of the Beam
Shear Diagram - Beam with 3 supports
- Thread starter teeth
- Start date
Homework Statement
Draw the shear diagram for the compound supported beam.
Homework Equations
Statics... then sum of moments at any betoken or forces in whatsoever direction are 0
The Attempt at a Solution
I summed the forces in the y direction, also as the moments at A and moments at B, but using these 3 equations causes the variables to abolish out when solving for the reactions. I go 0 = 0...
I estimate I cannot use two moment equations, but and so how am I to solve this? I'one thousand in first twelvemonth statics.
My equations are...
[tex]0 = \sum F_y = A_y + B_y + D_y -three*6 - 5[/tex]
[tex]0 = \sum M_A = B_y*3 + D_y*9 -3*half dozen*3 - 5*7.five[/tex]
[tex]0 = \sum M_B = -A_y*3 + D_y*6 - 5*4.5[/tex]
Counter-clockwise moment is positive
Answers and Replies
I tried getting isolating for [tex]C_y[/tex] simply I go along getting [tex]C_y = 0[/tex]
Is the upward force at pin C the same every bit the "shear strength" at C?
Sorry, is the bending moment at C nothing?
Yes. The pivot cannot develop a moment.
I tried getting isolating for [tex]C_y[/tex] but I keep getting [tex]C_y = 0[/tex]
Is the upward force at pin C the aforementioned every bit the "shear force" at C?
If y'all write equilibrium equations for beam C-D, you'll see that Cy cannot exist naught.
Homework Argument
Draw the shear diagram for the compound supported beam.
Homework Equations
Statics... then sum of moments at any bespeak or forces in any direction are 0
The Endeavour at a Solution
I summed the forces in the y direction, too equally the moments at A and moments at B, but using these iii equations causes the variables to cancel out when solving for the reactions. I go 0 = 0...
I guess I cannot use two moment equations, but and so how am I to solve this? I'one thousand in first year statics.My equations are...
[tex]0 = \sum F_y = A_y + B_y + D_y -3*6 - 5[/tex]
[tex]0 = \sum M_A = B_y*three + D_y*9 -3*6*3 - v*vii.5[/tex]
[tex]0 = \sum M_B = -A_y*3 + D_y*6 - 5*iv.five[/tex]
Counter-clockwise moment is positive
Hi Teeth, I had a query.
By taking the moments about Point A and B, You take formulated the equations 2 and iii.
Is this manner of generating equations right ?
Looking frontwards to a reply from you before long.
Aman Ratan
Cheers
Teeth hasn't been around PF since posting this question a year and a half agone, so I hope you lot are not looking for a reply any time soon.How-do-you-do Teeth, I had a query.By taking the moments virtually Point A and B, You have formulated the equations ii and 3.
Is this fashion of generating equations correct ?Looking forward to a answer from you soon.
Aman Ratan
Thank you
PF generally frowns on resurrecting old posts like this.
Information technology is OK to link to an quondam mail service if yous have a particular question about it, merely you should offset your own thread in that example.
Teeth hasn't been around PF since posting this question a twelvemonth and a half ago, and so I hope you are non looking for a reply any fourth dimension soon.PF by and large frowns on resurrecting quondam posts like this.
It is OK to link to an old post if yous take a particular question virtually it, but you should kickoff your own thread in that case.
Thanks SteamKing
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7-49 Draw the Shear and Moment Diagrams of the Beam
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