WORLD FOR CIVIL

Topic 7.2a: Secant Formula - Example 1

A steel I-beam (W 14 x 74) is used as a column. The beam is 20 ft. long and pinned on both ends. An eccentrically applied load of 280,000 lb. acts at the center of one flange as shown in the diagram. Young’s Modulus for steel is 30 x 10⁶ psi., and the yield stress for the steel is 40,000 psi. Let us first calculate the maximum compressive stress using the Secant Formula.

Beam Data

-	-	-	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-in2	d - in	wf - in	tf - in	tw - in	I - in4	S -in3	r - in	I - in4	S -in3	r - in
W 14x74	21.80	14.19	10.072	0.783	0.450	797.0	112.0	6.05	133.00	26.50	2.48

We replace the eccentric loading with an axial load plus a moment:
P =280,000 lb., M = P x e = 2,116,800 in-lb.

Next we calculate some constants:
Slenderness ratio = L/r = 20 ft x 12 in/ft / 6.05 in = 39.7
Eccentricity ratio = ec/r² = (7.56")(14.19"/2)/(6.05")² = 1.47
Axial Stress = P/A = 280,000 lb/ 21.8 in² = 12, 844 psi

Now we apply the secant formula:

or Maximum Compressive Stress = (12, 844 lb/in²)[ 1 + 1.47 sec {(39.7/2)*sqrt (12,844 lb/in² / 30x 10⁶ lb/in²)}]
= (12, 844 lb/in²)[ 1 + 1.47 (1.09)] = 33,440 lb/in²

We observe that the maximum compressive stress is less than the yield stress, 40,000 psi., thus this column is safe.

A second interesting question is what is the maximum load, P, which would result in a maximum compressive stress equal to the yield stress. For this we replace the maximum stress with the yield stress in the secant formula, and then solve the secant formula for the value of P.

Or 40,000 lb/in² = (P/21.8 in²)[ 1 + 1.47 sec{(39.7/2)sqrt(P/(30 x106 lb/in² * 21.8 in²))}]
Or simplifying a bit we have: 872,000 lb = P[1 + 1.47 sec {(19.85)sqrt(P/6.54 x 10⁸ lb)}]

This is a transcendental equation, and perhaps the most effective way to solve at this level is to guess solutions until we find an acceptable one.
We first try a value of P somewhat greater than the original. Let P = 350,000 lb.
Then we have: 872,000 lb. = 350,000 lb. [1 + 1.47 sec {(19.85)sqrt(350,000 lb./6.54 x 10⁸ lb.)}] or
872,000 lb. = 923,960 lb. (which is clearly not correct, our value of P is somewhat high, so we try again.

Let P = 330,000 lb., then 872,000 lb. = 330,000 lb. [1 + 1.47 sec {(19.85)sqrt(330,000 lb./6.54 x 10⁸ lb.)}],
or 872,000 lb. = 867,700 lb. Here we see we are quite close. So a load of 330,000 pounds will be very close to causing buckling to occur in the column.