If the diameter of a staybolt is doubled, how will its holding power be affected?

If the diameter of a staybolt is doubled, how will its holding power be affected?

• A. It will double
• B. It will quadruple
• C. It will halve
• D. It will remain the same

Answer: (B)

When considering the effect of doubling the diameter of a staybolt on its holding power, it is essential to recognize that the holding power is primarily related to the cross-sectional area of the bolt. The holding power, in terms of tensile strength, is proportional to the area through which the load is distributed. The cross-sectional area \( A \) of a circular bolt can be calculated using the formula: \[ A = \pi \times \left(\frac{d}{2}\right)^2 \] where \( d \) is the diameter of the staybolt. If the diameter is doubled, the new diameter becomes \( 2d \). The cross-sectional area with the new diameter can be expressed as: \[ A' = \pi \times \left(\frac{2d}{2}\right)^2 = \pi \times d^2 \] Now comparing this new area \( A' \) with the original area \( A \): 1. The original area is \( \pi \times \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \). 2. The new area is \( \pi \times (d)^2 \). Thus, when you double the

When considering the effect of doubling the diameter of a staybolt on its holding power, it is essential to recognize that the holding power is primarily related to the cross-sectional area of the bolt. The holding power, in terms of tensile strength, is proportional to the area through which the load is distributed.

The cross-sectional area ( A ) of a circular bolt can be calculated using the formula:

[ A = \pi \times \left(\frac{d}{2}\right)^2 ]

where ( d ) is the diameter of the staybolt. If the diameter is doubled, the new diameter becomes ( 2d ). The cross-sectional area with the new diameter can be expressed as:

[ A' = \pi \times \left(\frac{2d}{2}\right)^2 = \pi \times d^2 ]

Now comparing this new area ( A' ) with the original area ( A ):

1. The original area is ( \pi \times \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} ).

2. The new area is ( \pi \times (d)^2 ).

Thus, when you double the