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Here is a practise question that I am working on. The beam has different thickness due to the supports in the middle. I would expect the maximum bending stress be away from the middle due to the support? But that is not what I got.

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Also I am very confuse with the moving pointload. Do we have different induced maximum external moment (due to pointload) at different location along the beam(due to the thickness)? For example, the maximum moment at the thinner beam would be $(-1m \times 20kN)$? But the maximum moment at the thicker beam would be $(-2m \times 20kN)$? (As the pointload moves)

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What you have is a box beam with a different second moment of inertia at the segment without or with the cover plate, $I_1, I_2$, respectively. Note that the varying stiffness does not affect the distribution of shear force and moment along the span; so without guessing, you can compare the resulting bending stresses obtained from 1) place the concentrated loads at the mid-span, and 2) place the concentrated load at where the section change occurs (point "a").

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ADD: A cross-section is said to be symmetrical when the mirror image exists on the centroidal axis. In the graph below, the cross-section on the upperrow is said to be "doubly symmetrical", while the cross-sections in the lower row are said to be "singularity symmetrical about y-axis", for which, the equation for $I_x$ in the sketch is not applicable without modification.

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  • $\begingroup$ r13 this explanation and the diagrams are wonderful, thank you! So there will be different maximum moments cause by the point load for different thickness (as I hypothesised). And in essence this is analogous to sectioning. Following your advice, I calculated the maximum bending stress for the respective bending moment and second moment of area. Now I am unsure whether my calculations are correct because my maximum bending stress for the thick area is greater than the thin? Shouldn't the plates "strengthen" the middle beam and hence reduce the maximum bending stress? $\endgroup$
    – CountDOOKU
    Mar 6 at 4:05
  • $\begingroup$ Just to make sure the centroidal distance of the cross section with the plates is $\frac{0.122}{2}= 0.061m$? $\endgroup$
    – CountDOOKU
    Mar 6 at 4:30
  • $\begingroup$ @CountDOOKU r13's second sentence is very important. Note the r13 is basically saying that the bending moment diagram is not affected by the change in section. Also the maximum moment in a simply supported beam for a single point load occurs under the point load and the maximum moment in the beam will occur when the point load is at mid span. So step 1 is to calculate your applied moments. Step 2 is to convert your applied moments to stresses using the corresponding moment of inertias. $\endgroup$
    – Forward Ed
    Mar 6 at 4:59
  • $\begingroup$ @CountDOOKU since the plates are equal in size and the box is symentric, the centroid will be smack dab right in the middle of the section. So if you are using the outside edge of the plates as your reference point then yes the centroid is (110+2x6)/2 = 61 mm away from the outside face of the plates. $\endgroup$
    – Forward Ed
    Mar 6 at 5:01
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    $\begingroup$ @CountDOOKU You should review the "parallel-axis theorem for moments of inertia", for which you would need it all the time in your engineering career. I = Io + Ad^2. Also, due to the perfect symmetry of the tube and the cover plates, you can calculate I(2) = I(1) + I(pl). I(pl) is the moment of inertia of the rectangle that encloses the two plates subtract the moment of inertia of the area between the two plates. $\endgroup$
    – r13
    Mar 6 at 15:39

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