Question 3.17: The cantilever beam in Figure 3-24 is a steel American Stand...

Objective     Determine the maximum tensile and compressive stresses in the beam.
Given         The layout from Figure 3-24(a). Force = F = 10 000 lb; angle  θ= 30°. The beam shape: S6× 12.5; length = a = 24 in.

Section modulus = S = 7.37 in^3 ; area = A = 3.67 in^2 (Table A16-4).

TABLE A16-4 Properties of American Standard steel beams. S-shapes^*
				Flange		Axis X-X		Axis Y-Y
Designation	Area (in^2)	Depth (in)	Web thickness (in)	Width (in)	Average thickness (in)	I (in^4)	S (in^3)	I (in^4)	S (in^3)
S24× 90	26.5	24	0.625	7.125	0.87	2250	187	44.9	12.6
S20 × 96	28.2	20.3	0.8	7.2	0.92	1670	165	50.2	13.9
S20 × 75	22	20	0.635	6.385	0.795	1280	128	29.8	9.32
S20 × 66	19.4	20	0.505	6.255	0.795	1190	119	27.7	8.85
S18 × 70	20.6	18	0.711	6.251	0.691	926	103	24.1	7.72
S15 × 50	14.7	15	0.55	5.64	0.622	486	64.8	15.7	5.57
S12 × 50	14.7	12	0.687	5.477	0.659	305	50.8	15.7	5.74
S12 × 35	10.3	12	0.428	5.078	0.544	229	38.2	9.87	3.89
S10 X 35	10.3	10	0.594	4.944	0.491	147	29.4	8.36	3.38
S10 × 25.4	7.46	10	0.311	4.661	0.491	124	24.7	6.79	2.91
S8 × 23	6.77	8	0.441	4.171	0.426	64.9	16.2	4.31	2.07
S8 ×18.4	5.41	8	0.271	4.001	0.426	57.6	14.4	3.73	1.86
S7 × 20	5.88	7	0.45	3.86	0.392	42.4	12.1	3.17	1.64
S6 × 12.5	3.67	6	0.232	3.332	0.359	22 1	7.37	1.82	1.09
S5 × 10	2.94	5	0.214	3.004	0.326	12.3	4.92	1.22	0.809
S4 × 7.7	2.26	4	0.193	2.663	0.293	6.08	3.04	0.764	0.574
S3 × 5.7	1.67	3	0.17	2.33	0.26	2.52	1.68	0.455	0.39
*Data are taken from a variety of sources. Sizes listed represent a small sample of the sizes available. Notes: Example designation: S10 × 35 10 = nominal depth (in); 35 = weight per unit length (lb/ft) I = moment of inertia; S = section modulus.

Eccentricity of the load = e = 6.0 in from the neutral axis of the beam to the line of action of the horizontal component of the applied load.

Analysis      The analysis takes the following steps:
1.Resolve the applied force into its vertical and horizontal components.
2. Transfer the horizontal component to an equivalent loading at the neutral axis having a direct tensile force and a moment due to the eccentric placement of the force.
3. Prepare the free-body diagram using the techniques from Section 3-19.

4. Draw the shearing force and bending moment diagrams and determine where the maximum bending moment occurs.
5. Complete the stress analysis at that section, computing both the maximum tensile and maximum compressive stresses.

Results     The components of the applied force are:
F_x=Fcos(30°) = (10 000 \ lb)[cos(30°)] = 8660 lb acting to the right

F_y = Fsin(30°) = (10 000 \ lb)[sin(30° )] = 5000 lb acting downward

The horizontal force produces a counterclockwise concentrated moment at the right end of the beam with a magnitude of:

M_1 = F_x(6.0in) = (8660 \ lb) (6.0 \ in) = 51 960 lb.in

The free-body diagram of the beam is shown in Figure 3-24(b).
Figure 3-24(c) shows the shearing force and bending moment diagrams.
The maximum bending moment, 68 040 lb in, occurs at the left end of the beam where it is attached firmly to a column.
The bending moment, taken alone, produces a tensile stress (+) on the top surface at point B and a compressive stress ( – ) on the bottom surface at C. The magnitudes of these stresses are:

σ_1=± M/S = ± (68 040 \ lb \ in) / (7.37 in^3) = ± 9232 psi

Figure 3-24(d) shows the stress distribution due only to the bending stress. Now we compute the tensile stress due to the axial force of 8660 lb.

σ_2=F_x/A = (8660 \ lb)/(3.67 \ in^2) = 2360 psi

Figure 3-24(e) shows this stress distribution, uniform across the entire section. Next, let’s compute the combined stress at B on the top of the beam.

σ_B=+σ_1+σ_2= 9232 psi + 2360 psi = 11 592 psi Tensile

At C on the bottom of the beam, the stress is:

σ_C=-σ_1+σ_2= -9232 psi + 2360 psi = -6872 psi Compressive

Figure 3-24(f) shows the combined stress condition that exists on the cross section of the beam at its left end at the support. It is a superposition of the component stresses shown in Figure 3-24(d) and (e).