Analysis of a hyperstatic system by hand calculation 3

[DerGeraet]

Hi,

I'm currently trying to solve the problem, attached to the pdf-file. It's a hyperstatic system due to the fact that it's connected by two solid bearings (2 DOF*2). The forces in the horizontal direction, H1 and H7, are calculated but the vertical forces are kind of replaced? by K. Force F2 is in the middle of e1.A. I have really no idea, how to get the same results as listed on the file. I would be very thankful for every tip.

regards

 

[BAretired]

Had another look at this file. My previous expressions for rotation and deflection at Joint 1 with F only acting were wrong. They should be as follows: (see attachment)

(1) Force F acting upward at Joint 4
Simple span moment = F(L+2a)[sup]2[/sup]/4.
θ1 = -F(a[sup]2[/sup]+2ah+aL+L[sup]2[/sup]/4) (counterclockwise)
δ1 = Fh(aL+L[sup]2[/sup]/4+ah)/4 (left to right)

The expressions for H and M were correct before:
(2) Force H acting outward at Joints 1 and 7
M1 = M2 = 0
M3 = M4 = Hh
θ1 = +Hh(L+h)/2 (clockwise)
δ1 = -Hh2(L/2 + h/3) (right to left)

(3) Moment M at Joint 1 and -M at Joint 7
θ1 = +M(L/2 + h + a) (clockwise)
δ1 = -Mh(L+h)/2 (right to left)


1) -F(a[sup]2[/sup]+2ah+aL+L[sup]2[/sup]/4) + Hh(L+h)/2 + M(L/2+h+a) = 0
2) Fh(aL+L[sup]2[/sup]/4+ah)/4 - Hh2(L/2+h/3) - Mh(L+h)/2 = 0

Solving these two equations for M and H got too messy in terms of the variable names L,a and h, so I tried the values of 20, 5 and 10 for L, a and h.

The result was:
H = 0.3174F
M = 1.346F
which seems to pass the sanity check, so I think they are probably okay but I wouldn't guarantee it.



BA

 

[BAretired]

While it would be relatively easy to set up a spreadsheet to calculate the elastic solution of H and M for various values of L, a and h, it would be considerably easier to consider a plastic hinge forming at each corner of the bracket.

Mp = [φ]Z*F[sub]y[/sub] = [φ]wt[sup]2[/sup]F[sub]y[/sub]/4
where [φ] = 0.9
and Z, w and t are the plastic modulus, width and thickness of the bent plate.

Then F[sub]u[/sub]L/8 = Mp or F[sub]u[/sub]/2*a/2 = Mp
where F[sub]u[/sub] is the ultimate value of F.

So F[sub]u[/sub] = 8Mp/L or 4Mp/a, whichever is smaller.
If L>2a, 8Mp/L will govern.

Depending on which code one is using, F = F[sub]u[/sub]/LF where LF is the load factor (probably between 1.5 and 4, depending on the application).

BA

 

[BAretired]

To the first paragraph, I should have added "and at the load point". If L>2a, the failure mode would be a plastic hinge at Joints 3, 4 and 5. If L<2a, the failure mode would be a plastic hinge at Joints 1 and 2.

BA

 

[BAretired]

DerGaraet,
This is a link to a short spreadsheet I wrote for your bracket using OpenOffice.calc. It might be interesting to compare the results with the formula you provided in the original post.

BA