WORLD FOR CIVIL

STATICS & STRENGTH OF MATERIALS - Example

A loaded, simply supported beam is shown below. For this beam:

Select the best T-beam to use if:
1). The maximum allowable bending stress = 30,000 lb/in².
2). The maximum allowable shear stress = 12,000 lb/in².
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.

1.) FBD of structure (See Diagram)
2.) Resolve all forces into x/y components
3.) Apply equilibrium conditions:

Sum F_x = 0 none
Sum F_y = A_y + C_y - 5,000 lbs - 1,000 lbs/ft (8 ft)= 0
Sum T_A = -5,000 lbs (4 ft) + C_y (8 ft) - 1,000 lbs/ft (8 ft)(12 ft)= 0
Solving: A_y = -1500 lbs; C_y = 14,500 lbs

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 = -32,000 ft-lb.; and V_max = 8,000 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 = 30,000 psi.; and Maximum Allowable Shear Stress = 12,000 psi.

We now use the flexure formula form: 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:
30,000 lb/in² = (32,000 ft-lb.)(12 in./ft.)/ S; and then S = 384,000 in-lb./ 30,000 lb/in² = 12.8 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. Shown below is a selection of beams. We would like to now selected the best beam based on the minimum value of the section modulus determined above. We select the beam but 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).

Step 4. After examining the selections, we determine WT 12 x 34 is the best beam from the selection listed. It has a section modulus of 15.6 in³ (greater than the minimum section modulus of 12.8 in³), and a weight of 34 lb./ft, which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.

T-Beam Data

-	-	Depth	Flange	Flange	Stem	-	Cross	Section	Info.	-
Designation	Area	of T	Width	thick	thick	-	x-x axis	x-x axis	x-x axis	x-x axis
-	A	d	bf	tf	tw	d/tw	I	S	r	y
-	in2	in	in	in	in	-	in4	in3	in	in
WT 12x34	10.00	11.86	8.961	0.582	0.416	28.50	137.00	15.600	3.700	3.070

Step 5: Now using the selected beam we check that the maximum horizontal shear stress (HSS) in the beam is within the allowable stress.
t = Vay' / Ib = (8000 lb.)(8.83" * .42")(4.42") / (137 in⁴ * .42") = 2280 lb/in² which is lower than the allowable shear stress of 12,000 lb/in². So the beam is safe and is our best choice of the beams given to choose from.