Help with Unit Load Method for Gable Frame

Question

Help with Unit Load Method for Gable Frame

Tygra_1983 (Student)

(OP)

24 Nov 23 14:50

Hi guys,

Back again!

I am attempting to solve a pinned support gable frame using the unit load method.

I am using BeamGuru software, but I want to check the results to esure the program is working properly.

I have replaced the pinned supporta at A with a roller and it looks like this:

The height to eaves is 8 metres, the height to the apex is 9.58 metres. The frame is 30 metres long and the roof is at an angle of 6 degrees.

I have been using Octave to perform the integrations:

CODE -->

syms x E I
      % Real structure
      Mxr1 = (196.787*cosd(6) - 0.587*sind(6)).*x - 13.14.*x.^2/2 % left rafter
      Mxr2 = (197.413*cosd(6) + 0.587*sind(6) - 1.174*sind(6)).*x + 9.392 - 13.14.*x.^2/2; % Right rafter
      Mxc2 = -1.174*x  % right-hand column

      % Virtual structure
      mxr1 = -sind(6)*x + 8   % Left rafter
      mxr2 = sind(6)*x + 8    % Right rafter
      mxc2 = -x               % Right-hand column

       % Integrations over excah member length for real and virtual structures
             Ix1 = 1/(E*I)*int(Mxr1*mxr1,15.083,0)
             Ix2 = 1/(E*I)*int(Mxr2*mxr2,15.083,0)
             Ix3 = 1/(E*I)*int(Mxc2*mxc2,8,0)
             ix1 = 1/(E*I)*int(mxr1.^2,15.083,0)
             ix2 = 1/(E*I)*int(mxr2.^2,15.083,0)
             ix3 = 1/(E*I)*int(mxc2.^2,8,0)
      
       % calculate the horizontal reaction at A
     HA = (Ix1+Ix2+Ix3)/(ix1+ix2+ix3)

For HA, from these calculations I am getting 111.90 kN. The software is giving me 98.6 kN. So, Im a bit off. I think I could be getting mixed up with signs, but I have checked and checked, and cannot see why I am getting the incorrect result.

Many Thanks!

rb1957 (Aerospace) · Answer 1 · 2023-11-24 15:04:11

in your virtual model, should "8" be "8.79" (the average of the roof heights) ? ... 10% difference ??

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Tygra_1983 (Student) · Answer 2 · 2023-11-24 19:31:52

I have always understood it that you take the moment at the start of the section, which is 8 kNm.

If you do this you get:

-sin(6)*x - 8

So, the moment at the end of the member (the apex) is:

-0.105*15.083 - 8 = -9.58 kNm ---> this is the moment at the apex.

rb1957 (Aerospace) · Answer 3 · 2023-11-24 19:34:50

no, your moment is from the center of effort of the force ... you're replacing the distributed force with a point load, through the center of the distributed force, yes?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Tygra_1983 (Student) · Answer 4 · 2023-11-24 19:48:58

-sin(6)*x - 8 is the general equation for the moment along the member. Plug in any x value and you will get the moment at that particular point.

rb1957 (Aerospace) · Answer 5 · 2023-11-24 20:15:30

ok, what is "mxr1" ? sind(6)*x is the height about the lower edge, so "+8" is the height about ground. ok, but why "-" of the left and "+" on the right ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

BAretired (Structural) · Answer 6 · 2023-11-25 02:54:55

If you just want to check your program, you could use Kleinlogel Rigid Frame Formulas (see below).

Tygra_1983 (Student) · Answer 7 · 2023-11-25 13:32:49

Hi BAretired,

Thing is my frame has equivalent horizontal forces applied to it. Not just vertical loads.

Tygra_1983 (Student) · Answer 8 · 2023-11-25 14:17:47

Its okay now guys, I have sussed it. I wasn't including the virtual loads on the left-hand column. Because for the real structure the loads were zero for the left-hand column, it escaped me that there were indeed loads for the virtual.

rb1957 (Aerospace) · Answer 9 · 2023-11-25 18:17:06

good on you for figuring it out. thx for closing the thread.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Tygra_1983 (Student) · Answer 10 · 2023-11-25 19:57:08

RE: Help with Unit Load Method for Gable Frame

Tygra_1983 (Student)

(OP)

25 Nov 23 19:57

Thanks rb1957.

BAretired (Structural) · Answer 11 · 2023-11-26 21:59:16

Quote (Tygra 1983)
Hi BAretired,

Thing is my frame has equivalent horizontal forces applied to it. Not just vertical loads.
Gre

I too am pleased that you were able to resolve the problem on your own Tygra, but for those who may believe that Kleinlogel Rigid Frame Formulas deal exclusively with vertical loads, please be aware that the tables deal with any type of load, including horizontal loads. There are several pages associated with Frame 89, which I did not post.

Prior to the computer age, structural engineers regarded Kleinlogel as the go-to reference for rigid frame design.

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Help with Unit Load Method for Gable Frame