Shear stress resisted by a bolt

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fonseh
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Homework Statement


In this problem , why the author use (30x10^3) /s = VQ / I only ? Shouldnt it be 4(30x10^3) /s = VQ / I

Homework Equations

The Attempt at a Solution


I said so , because the board at the centered is connected by 4 bolts , 2 on the left and 2 on the right . The faces on the right also resisted by the same arrangement of nails , right ?
 
Attachments
  • Textbook figure of a built-up wooden beam made of three boards bolted together, with a downward shear force V labeled on the center board and dimensions of 0.5 in., 1 in., 4 in. and 3 in. marked on the side.
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  • Handwritten solution to problem F7-10: I = (1/12)(1)(6^3) + 4[(1/12)(0.5)(4^3) + 0.5(4)(3^2)] = 100.67 in^4; Q = y'A' = 3(4)(0.5) = 6 in^3; q_allow = F/s = 6/s; q_allow = VQ/I = 15(6)/100.67; s = 6.711 in.; use s = 6 in. 8.
    520.PNG
    6.8 KB · Views: 523
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fonseh said:

Homework Statement


In this problem , why the author use (30x10^3) /s = VQ / I only ? Shouldnt it be 4(30x10^3) /s = VQ / I

Homework Equations

The Attempt at a Solution


I said so , because the board at the centered is connected by 4 bolts , 2 on the left and 2 on the right . The faces on the right also resisted by the same arrangement of nails , right ?
Where did the 30x10^3 come from? Regardless, your method is wrong. In determining Q for bolt shear, either you use the top right board full area, or the top left board full area, or the middle and 2 bottom areas divided by 2 ( this last method not recommended, but may help conceptually with your understanding).