WORLD FOR CIVIL

Topic 5.4a: Beam Selection - Example 1

A loaded, cantilever beam is shown in Diagram 1. We would like to choose the best T-beam to use from the T-Beam selection shown at the lower part of this page.
The maximum allowable bending stress = 35,000 lb./in2. (both in tension and compression), and the maximum allowable shear stress = 15,000 lb/in2 for the beam material used.

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 - (1,500 lb./ft.)(6 ft) - (1,000 lb./ft.)(2 ft) - (800 lb./ft.)(4 ft) = 0
Sum T_B = -(1,500 lb./ft.)(6 ft)(3 ft.)-(1,000 lb./ft.)(2 ft)(7 ft.)-(800 lb./ft.)(4 ft)(10 ft.) + M_ext = 0
Solving: A_y = 14,200 lb.; M_ext = 73,000 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 = -73,000 ft-lb.; and V_max = 14,200 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 = 15,000 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:
35,000 lb/in² = (73,000 ft-lb.)(12 in./ft.)/ S; and then S = 876,000 in-lb./ 35,000 lb/in² = 25 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).

After examining the selections, we determine WT 15 x 49.5 is the best beam from the selection listed. It has a section modulus of 30.1 in³ (greater than the minimum section modulus of 25 in³), and a weight of 49.5 lb./ft, which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.

-	-	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 7 x157	46.20	8.60	16.235	2.283	1.415	6.07	179.00	27.000	1.970	1.980
WT 7x143.5	42.20	8.41	16.130	2.093	1.310	6.42	157.00	24.100	1.930	1.870
WT 7x132	38.80	8.25	16.025	1.938	1.205	6.85	139.00	21.500	1.890	1.780
WT 7x123	36.20	8.13	15.945	1.813	1.125	7.22	126.00	19.600	1.860	1.710
WT 13.5x57	16.80	13.64	10.070	0.932	0.570	23.90	289.00	28.300	4.150	3.410
WT 13.5x51	15.00	13.54	10.018	0.827	0.518	26.10	258.00	25.400	4.140	3.380
WT 13.5x47	13.80	13.46	9.990	0.747	0.490	27.50	239.00	23.800	4.150	3.410
WT 13.5 x42	12.40	13.35	9.963	0.636	0.463	28.80	216.00	22.000	4.180	3.500
WT 15x66	19.40	15.15	10.551	1.000	0.615	24.60	421.00	37.400	4.650	3.900
WT 15x62	18.20	15.08	10.521	0.930	0.585	25.80	395.00	35.300	4.650	3.890
WT 15x58	17.10	15.00	10.500	0.850	0.564	26.60	372.00	33.600	4.670	3.930
WT 15x54	15.90	14.91	10.484	0.760	0.548	27.20	350.00	32.100	4.690	4.020
WT 15x49.5	14.60	14.82	10.458	0.670	0.522	28.40	323.00	30.100	4.710	4.100

Step 4. The last step is to now check that the beam we have selected is also safe with respect to the horizontal shear stress – that is, that the maximum horizontal shear stress for the selected beam is within the allowable shear stress for the beam material. We therefore now apply our formula for the horizontal shear stress for the WT 15 x 49.5 T-beam: Shear Stress = Vay'/Ib
The maximum shear stress occurs at the neutral axis of the beam:
V = maximum shear force = 14,200 lb. (from the shear force diagram)
I = moment of inertia of cross section, from beam table; I = 323 in⁴.
b = width of beam where we wish to find shear stress (neutral axis for maximum) from table; b = .522 in.
a = area from point we wish to find shear stress at (neutral axis) to an outer edge of beam. In this case we will go to bottom of beam. Then a = (.522" * 10.72" )= 5.6 in².
y' = distance from neutral axis to the centroid of the area "a" ; y' = 5.36 in.
Maximum Horizontal Shear Stress = Vay'/Ib = (14,200 lb)*(5.6 in²)*(5.36 in)/ (323 in⁴)(.522 in) = 2530 lb/in²

We see that this value is well within the allowable shear stress of 15,000 lb/in2 given above. Thus we have selected the best beam to use from the given list of possible beam.