F=32KN; Mf1=60KN-m; Mf2=-58KN-m; g=30KN/m I need to solve this using Force method. Find the reaction at the pinned end.

  • $\begingroup$ It would be great if anyone could help. I have tried finding the reaction force, but I'm getting it wrong. The correct answer is 2.6KN $\endgroup$
    – Ravi
    Mar 24 '19 at 7:29

Your bending moment diagram is very wrong. It is continuous even though there are concentrated bending moments (which create discontinuities), and you have $F$ generating positive bending moment when it should be negative.

Using your notation, where subscript:

  • 1 represents the pinned end;
  • 2, where $F$ and $M_{f1}$ are applied;
  • 3, where $M_{f2}$ is applied;
  • 4, the fixed end

and adding that superscript - means the result on the left side of the point and +, on the right side of the point, we can calculate that:

$$\begin{alignat}{4} M_1 &= &&=0\text{ kNm} \\ M_2^- &= &&=0\text{ kNm} \\ M_2^+ &= M_2^- - 60 &&= -60\text{ kNm} \\ M_3^- &= -60 - 32\cdot3.6 &&= -175.2\text{ kNm} \\ M_3^+ &= M_3^- + 58 &&= -117.2\text{ kNm} \\ M_4 &= -60 - 32\cdot6.4 + 58 - 30\cdot2.8\cdot\dfrac{2.8}{2} &&= -324.4\text{ kNm} \\ \end{alignat}$$

Running this on Ftool to confirm:

enter image description here

You seem to be using the moment-area method to calculate deflections. As far as I can tell, you've done so correctly, just using the wrong moment diagram. Just repeat your work and it should be fine.

To compare your results, here are the ones given by Ftool $(EI = 1\text{ kNm}^2)$:

  • Total deflection at the "free end": 5812 m
  • Deflection due to moment-area 1: 820.8 m
  • Deflection due to moment-area 2: 912.4 m
  • Deflection due to moment-area 3: 2297 m
  • Deflection due to moment-area 4: 1782 m

where moment-area 1 is the rectangular component of the 3.6 m stretch, area 2 is the inclined component of the 3.6 m stretch, area 3 is the rectangular component of the 2.8 m stretch, and area 4 is the parabolic component of the 2.8 m stretch.

  • 2
    $\begingroup$ A minor point of interest is that in the United States it’s fairly common to choose a convention where positive moments are drawn on the compression side of the member (particularly for steel design). $\endgroup$
    – CableStay
    Apr 3 '19 at 3:37
  • $\begingroup$ @CableStay, learn something new every day! I'll update my question when I get to my computer. $\endgroup$
    – Wasabi
    Apr 3 '19 at 10:24
  • $\begingroup$ @Wasabi thanks for the answer... though, when i did the calculations with moment-area method, the answer again turns out to be wrong.. Maybe the answer set is wrong... $\endgroup$
    – Ravi
    Apr 4 '19 at 13:54
  • $\begingroup$ @Ravi: I've updated my answer with the deflection at the free end (assuming $EI=1$), as well as all the partials. Check your math against that to see where you've gone wrong. $\endgroup$
    – Wasabi
    Apr 5 '19 at 0:55

The steps to find the reaction at the pin support using conventional methods:

  1. Remove the redundant support at the pin to create a statically determinate virtual structure.
  2. Solve for the cantilever tip support deflection under the provided loading. Use principle of linear superposition to add up the deflection from each load source separately.
  3. In order to make the virtual structure compatible with the real structure, a point load at the tip is required to counter act the deflection of the virtual structure.

It looks like you went through these steps. My guess is there is a problem with the deflection computation. I would try breaking apart each load individually and finding the deflection for each. Then add them up. Then enforce compatibility.


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