Strength of Materials Review Problem

[LFRIII]

Going back to strength of materials book to refresh after many years. Working through problems. Looking for help solving the following problem:

2. A cantilever beam is composed of two 6x6-in timbers held together by bolts and connector rings. The beam is 3 1/2 feet long as has a concentrated load applied at the end. The bolt holes are 3/4" in diameter and each connector ring can safely transmit a force of 6000 lb. in shear. If the load P=5000 lb., what is the required spacing of the bolts? Answer e=9.6 in.

Got that one. Next one is the mystery

3. Calculate the maximum bending stress in the beam of the preceding problem assuming that the first bolt on the left is 2 in. from the left wall. Answer sigma = 1590 psi.

Any assistance would be appreciated.

 

[le99]

Q2 can be solved by VQ/I; Q3 is simply My/I (assuming the beam is fully composite).

 

[BAretired]

Are we assuming 6"x6" is the actual dimension of the beams? If so, the numbers for Q3 are close but not 1590psi. But to take them as a more realistic 5.5"x5.5", it's even worse.

What is the significance of the first bolt being 2" from the wall? Does that enter into the calculation?

BA

 

[Ingenuity]

What is the significance of the first bolt being 2" from the wall? Does that enter into the calculation?

Critical bending stress would be based on a reduced cross section where the 3/4" hole is deducted from the gross 12"D x6"W section.

I[sub]total[/sub] = 12[sup]3[/sup]*6/12 = 864 in[sup]4[/sup]

So I[sub]nett[/sub] = 12[sup]3[/sup]*(6-0.75)/12 = 756 in[sup]4[/sup]

y = 6"

Maximum moment at face of cantilever: 5 kips * 42" = 210 k.in ==> σ = 210 k.in * 6 / 864 = 1458 psi on total section

Moment at first bolt hole: 5 kips *40" = 200 k.in ==> σ = 200 k.in * 6 / 756 = 1587 psi <== GOVERNS

 

[BAretired]

Okay, hadn't thought about doing that, but I agree with Ingenuity.

BA