WORLD FOR CIVIL

Topic Exam Solution - Problem 2

A loaded, simply supported beam is shown below. For this beam select the best I-beam to use if: the maximum allowable bending stress = 35,000 lb./in²(tension & compression), and the maximum allowable shear stress = 10,000 lb./in².

Solution:

STEP 1: Apply Static Equilibrium Principles and determine the external support reactions:
1.) FBD of structure (See Diagram 1)
2.) Resolve all forces into x/y components
3.) Apply equilibrium conditions:
Sum F_x =A_x = 0
Sum F_y = A_y - (800 lb./ft.)(8 ft) - (1,200 lb./ft.)(8 ft) + = 0
Sum T_A = -(800 lb./ft.)(8 ft)(4 ft.)-(1,200 lb./ft.)(8 ft)(12 ft.) +C_y(12 ft.) = 0
Solving: A_y = 4,270 lb.; C_y = 11,730 ft-lb.

STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the Beam, and determine the values of the maximum bending moment and maximum shear force. (See Diagrams 2 and 3.)
From the Diagrams we observe that M_max = 11,400 ft-lb.; and V_max = -6,930 lb.

Step 3: Use the Flexure Formula for maximum bending stress and the specifications for the beam material to determine the minimum Section Modulus needed to carry the load. By material specification we mean the allowable stresses (tensile, compressive, and shear) for the beam material. This information is normally furnished by the beam supplier with their selection of beams. The allowable stresses were given at the beginning of this problem as: Maximum Allowable Bending Stress = 35,000 psi.; and Maximum Allowable Shear Stress = 10,000 psi.

We now use the flexure formula form: Maximum Bending Stress = M_max / S, and use the lowest allowable axial stress for the maximum bending stress, and solve for the value of the section modulus. Placing values into the equation we have:
35,000 lb./in² = (11,400 ft-lb.)(12 in./ft.)/ S; and then S = 136,800 in-lb./ 35,000 lb./in² = 3.91 in³ .

This value for the section modulus is the smallest value possible if the maximum bending stress is not to exceed the allowable axial stress for the beam material. We would like to now selected the best beam based on the minimum value of the section modulus determined above. We select the beam from the beam table - and find the one with a section modulus equal or greater than the minimum section modulus and with the least pounds per foot weight (which normally means the least expensive beam).

After examining the selections, we determine W 5 x 16 is the best beam from the selection listed. It has a section modulus of 8.5 in³ (greater than the minimum section modulus of 3.91 in³), and a weight of 16 lb./ft, which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.

Step 4. We now must check the maximum shear stress in the I-Beam to see if it is below the allowable shear stress. We can use the approximate formula for I-beam, Maximum Horizontal Shear Stress = V_max/A_web. This says the approximate maximum shear stress in an I - Beam is equal to the maximum shear force divided by the area of the web of the I-Beam. Applying this we have:

V_max = maximum shear force = 6,930 lb. (from the shear force diagram)
A_max = area of web = (.24" * 4.28" )= 1.03 in² (from the beam table for the W 5 x 16 beam.)

Then Maximum Horizontal Shear Stress = (6,930 lb.)/(1.03 in²) = 6730 lb./in² As long as this approximate value is reasonably below the allowable shear stress for the beam material there is no need to use the exact formula for the maximum shear stress. Please remember, however, the approximate formula is only for the maximum horizontal shear stress (which occurs are the neutral axis) in an I-Beam. If we need to know the shear stress at any other location, we must use the standard formula.
We see that this value is well within the allowable shear stress of 10,000 lb./in2 given above. Thus we have selected the best beam to use from the given list of possible beam.