What is the Allowable Load on a Tie-Down with a Bent Bar and Fiberglass Support?

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Homework Help Overview

The discussion revolves around determining the allowable load on a tie-down system involving a bent bar and fiberglass support. The problem includes specific parameters such as forces, dimensions, and material properties related to shear stress and bearing pressure.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of allowable loads based on shear stress and bearing pressure, referencing specific formulas for area and load. Questions arise regarding the understanding of shear stress calculations.

Discussion Status

Some participants have provided calculations and expressed understanding of the methods used, while others are seeking clarification on specific concepts, indicating an ongoing exploration of the topic.

Contextual Notes

Participants are working within the constraints of allowable stress values and specific dimensions provided in the problem statement. There is a noted lack of consensus on the understanding of shear stress calculations.

Dyls
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okay, I've been banging my head against a wall for a few hours. It's a bit difficult to describe but here goes...

There is a rope being pulled up with force P. The bottom of this rope is tied to a bent bar. The bent bar is shaped like an upside down U with the two ends stuck in a piece of fiberglass that has a thickness of t=(3/8) in. After the plywood on each "prong" is a washer and a screw-head. The screw going into the plywood has a diameter of 1/4 in and the washer has a diameter of 7/8 in. Allowable shear stress in the fiberglass is 300 osi and the allowable bearing pressure between the washer and the fiberglass is 550 psi. What is the allowable load on the tie-down?

The answer is 607 lb and 619 lb but I can't seem to figure how the book got this. Can anyone help me?
 
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[tex]550~psi = \frac {L}{A_{washer}}[/tex]

[tex]A_{washer} = \frac{\pi}{4} (D^2_{out} - D^2 _{in})[/tex]

Allowable load = 2L = 607.46 lbs

Similarly for the second part, keeping in mind that the relevant area, A for the shear stress is

[tex]A = \pi D_{out} \cdot t[/tex]

This will give you load = 618.50 lbs

Do you understand why this is so ?
 
wow... you rock. Yeah I understand why. Thanks so much!
 
Well, actually I don't quite understand the shear stress part. Can you explain that?
 

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