WORLD FOR CIVIL

Topic 5.4: Beams -Beam Selection

Another very important use of the flexure formula is in Beam Selection. That is, how does one decide on the best (safe and least expensive) beam to use with a particular loading. Perhaps the best way to explain this process is to work carefully through an example of the procedure.

In Diagram 1 we have shown a loaded beam (same loading as in Beams II - Bending Stress Example II). For this beam and loading we would like to select the best I-Beam to use (from a selection of I-Beams, which we will discuss shortly)

STEP 1: Apply Static Equilibrium Principles and determine the external support reactions.
1.) Draw Free Body Diagram of structure (See Diagram 2)
2.) Resolve all forces into x & y components
3.) Apply equilibrium conditions:
Sum F_x = 0 none
Sum F_y = B_y + D_y - 2,000 lbs/ft (4 ft) - 5,000 lbs = 0
Sum T_B = 5,000 lbs (4 ft) - 2,000 lbs/ft (4 ft) (6 ft) + D_y(8 ft) = 0
Solving: B_y = 9,500 lbs; D_y = 3,500 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. The graphs are shown in Diagram 3 and Diagram 4. (We determine these diagrams in some detail in Topic 4.8a: Bending Stress - Example 2. If you need more detail please see that example.)

From the Diagrams we observe that M_max = -20,000 ft-lb.; and V_max = -5000 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. For this example we will use the following allowable stresses for the beam material:

; ;

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:

20,000 lb/in² = (20,000 ft-lb.)(12 in./ft.)/ S; and then S = 240,000 in-lb./ 20,000 lb/in² = 12 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 W 8 x 17 is the best beam from the selection listed. It has a section modulus of 14.1 in³ (greater than the minimum section modulus of 12 in³), and a weight of 17 lb/ft, which is the least weight for beams with a section modulus greater than the minimum from the beam selection listed below.

-	-	-	Flange	Flange	Web	Cross	Section	Info.	Cross	Section	Info.
Designation	Area	Depth	Width	thick	thick	x-x axis	x-x axis	x-x axis	y-y axis	y-y axis	y-y axis
-	A	d	bf	tf	tw	I	S	r	I	S	r
-	in2	in	in	in	in	in4	in3	in	in4	in3	in
W 6x25	7.35	6.37	6.080	0.456	0.320	53.3	16.7	2.69	17.10	5.62	1.53
W 6x20	5.88	6.20	6.018	0.367	0.258	41.5	13.4	2.66	13.30	4.43	1.51
W 6x15.5	4.56	6.00	5.995	0.269	0.235	30.1	10.0	2.57	9.67	3.23	1.46
W 8x20	5.89	8.14	5.268	0.378	0.248	69.4	17.0	3.43	9.22	3.50	1.25
W 8x17	5.01	8.00	5.250	0.308	0.230	56.6	14.1	3.36	7.44	2.83	1.22
W 8x67	19.70	9.00	8.287	0.933	0.575	272.0	60.4	3.71	88.60	21.40	2.12
W 8x40	11.80	8.25	8.077	0.558	0.365	146.0	35.5	3.53	49.00	12.10	2.04
W 8x35	10.30	8.12	8.027	0.493	0.315	126.0	31.1	3.50	42.50	10.60	2.03
W 8x31	9.12	8.00	8.000	0.433	0.288	110.0	27.4	3.47	37.00	9.24	2.01

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 maximum horizontal shear stress in an I-Beam: _max = V_max/A_web = 5000 lb/ (.230 x 7.384) = 2944 lb/in². We see that this value is well within the allowable shear stress of 16,000 lb/in2 given above. Thus we have selected the best beam to use from the given list of possible beam.

We now look at an example selecting the best T-Beam to use for a particular loading.
Please Select Topic 5.4a: Beam Selection Example 1

When finished with Example of Beam Selection,