Question

A steel angle has a cross-section as indicated. Find the horizontal neutral axis (N.A.) and moment of inertia (l) for the indicated shape. 0.57" part A& Ay Ad2 4.00

1. A steel angle has a cross-section as indicated. Find the horizontal neutral axis (N.A.) and moment of inertia (I) for the indicated shape.

2. The steel angle from problem 1 is used as a beam to span 8.5 feet. A heavy mechanical pipe is suspended from the beam at mid­span. The load from the pipe is 2,350 lbs.

a)   Draw a free body diagram for the beam, and determine the reactions at each end of the beam.

b)   The moment produced by the load is 4,990 ft-lbs. Detemine the maximum bending stress in the angle.

c)   Which side of the angle (top or bottom) has tensile stress?

Please show work! Thank you!

Answer 1

As the beam is uniform and there is no curve on the beam before bend. The neutral axis will be at the centroid of the body.

Taking the corner of the L shaped beam as origin, and x and y direction as shown in the figure.

0.57 6.26" 1 C1,663, 2. 2 8 3) 2. 0.74 (o,o) 2 KL"

a) x coordinate of centroid:

x = (A₁x₁ + A₂x₂) / (A₁ + A₂)

A₁ = 6.26 x 0.57 = 3.568 inch²

A₂ = 4 x 0.74 = 2.96 inch²

x₁ = 0.57/2 = 0.285''

x₂ = 4/2 = 2''

x = (3.568x0.285 + 2.96x2) / (3.568 + 2.96)

x = 1.063''

b) y coordinate of centroid:

y = (A₁y₁ + A₂y₂) / (A₁ + A₂)

y₁= (6.26/2) + 0.74 = 3.87''

y₂= 0.74/2 = 0.37''

y = (3.568x3.87 + 2.96x0.37) / (3.568 + 2.96)

y = 2.283''

Therefore, the neutral axis will pass through the point (1.063, 2.283).

Moment of Inertia (MOI) of the L shaped beam about the horizontal axis ''X - X'' passing through the point (1.063, 2.283):

MOI of body about an axis parallel to its centroid is given as:

(MOI about centroid) + (Area x distance between both axes)

MOI of part 1 = (1/12)b₁d₁³ + A₁r₁² r₁ = |y₁ - y|

For part 1;

(MOI)₁ = (1/12)(0.57x6.26³) + (3.568)(3.87-1.063)²

(MOI)₁ = 39.766 inch⁴

(MOI)₂ = (1/12)b₂d₂³ + A₂r₂² r₂ = |y - y₂|

(MOI)₂ = (1/12)(4x0.74³) + (2.96)(2.283-0.37)²

(MOI)₂ = 10.967 inch⁴

MOI of whole cross section, I = (MOI)₁ + (MOI)₂

I = 39.766 + 10.967

I = 50.733 inch⁴

Now, a heavy mechanical pipe is suspended from the beam at mid­span and the load from the pipe is 2,350 lbs.

The beam is supported at the ends, let the reaction forces be R₁ and R₂. Because of symmetry, the load will be divided equally at both the supports and the reaction will be same and equal to;

R₁ = R₂ = 2350/2 = 1175 lbs (As shown in the Free Body Diagram below)

= 1175 lbs 2. 1 17.5 lbs 23so lbs 2 3 soRbs 175 Lbs 1175 lbs

It is given that Moment on the beam is 4990 ft-lbs.

Bending stress in the beam is given by;

σ = My / I

σ is maximum for maximum value of y, where y = distance of point (at which σ is calculated) from Neutral axis.

Maximum value of y is for the top most point of part 1.

y = 7 - 2.283 = 4.717'' or 0.393 ft

I = 50.733 inch⁴ = 2.45x10^-3 ft⁴

σ = (4990x0.393) / (2.45x10^-3)

σ = 800436.7 lbs/ft²

Compression Ten sion.

Due to loading, the beam will bend to form an arc. The bottom fibers of the beam will experience tensile forces and top part of the beam will be in compression. Note that the Bending stress is maximum at the top fibers and is compressive in nature.