5. Two pieces of steel are held together with six bolts. One end of the first piece of steel is welded to a beam and the second piece of steel is bolted to the first. There is a load of 54,000 pounds applied to the end of the second piece of steel. The tensile strength of steel is 74,000 psi, the shear strength is 48,000 psi.

a) Calculate the diameter of the bolts needed to support the load.

b) Find the width of the steel if the thickness of the steel is 1/4”. The bolt holes are 1/16” larger than the bolt.

Two pieces of steel are held together with six bolts. One end of the first piece of steel is welded to a beam and the second piece of steel is bolted to the first. There is a load of 54,000 pounds applied to the end of the second piece of steel. The tensile strength of steel is 74,000 psi, the shear strength is 48,000 psi.

a) Calculate the diameter of the bolts needed to support the load.

b) Find the width of the steel if the thickness of the steel is ¼”. The bolt holes are 1/16” larger than the bolt.

To calculate the diameter of the bolts needed to support the load, we need to consider both the tensile and shear strength of the bolts.

a) Tensile Strength:
The load applied to the second piece of steel is 54,000 pounds. Since the tensile strength of the steel is given as 74,000 psi (pounds per square inch), we can divide the load by the tensile strength to determine the minimum cross-sectional area required to support the load in tension.

Minimum cross-sectional area = Load / Tensile Strength
= 54,000 pounds / 74,000 psi
= 0.729 square inches

Since we have six bolts, we can divide this area by six to get the minimum cross-sectional area required for each bolt.

Minimum cross-sectional area per bolt = 0.729 square inches / 6
= 0.122 square inches

Now, we need to find the diameter of a bolt with a cross-sectional area of 0.122 square inches.

To calculate the diameter, we can use the formula for the area of a circle:

Area of a circle = π * (diameter/2)^2

Rearranging the formula, we get:

diameter = sqrt((4 * Area of a circle) / π)

Substituting the values, we have:

diameter = sqrt((4 * 0.122 square inches) / π)
≈ sqrt(0.488 square inches / π)
≈ sqrt(0.1553)
≈ 0.394 inches

So, the diameter of the bolts needed to support the load is approximately 0.394 inches.

b) Width of the steel:
To find the width of the steel, we need to take into account the thickness of the steel, the size of the bolt holes, and the number of bolts used.

The thickness of the steel is given as 1/4 inch. Since the bolt holes are 1/16 inch larger than the bolt, the effective diameter of each bolt hole is:

Effective diameter of each bolt hole = bolt diameter + bolt hole size increase
= 0.394 inches + 1/16 inch ≈ 0.438 inches

Since there are six bolts, the total width of the steel would be the sum of the bolt hole diameters plus the thickness of the steel.

Total width = (number of bolts * effective diameter of each bolt hole) + thickness of the steel
= (6 * 0.438 inches) + 1/4 inch
= 2.628 inches + 1/4 inch
= 2.878 inches

Therefore, the width of the steel would be approximately 2.878 inches.