# Mechanics of Materials

1. Jan 16, 2005

### Dyls

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?

2. Jan 16, 2005

### Gokul43201

Staff Emeritus
$$550~psi = \frac {L}{A_{washer}}$$

$$A_{washer} = \frac{\pi}{4} (D^2_{out} - D^2 _{in})$$

Allowable load = 2L = 607.46 lbs

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

$$A = \pi D_{out} \cdot t$$

This will give you load = 618.50 lbs

Do you understand why this is so ?

3. Jan 16, 2005

### Dyls

wow... you rock. Yeah I understand why. Thanks so much!

4. Jan 16, 2005

### Dyls

Well, actually I don't quite understand the shear stress part. Can you explain that?