Question 10.2: For the system shown in Figure 10–15, determine the maximum ...

Objective    Compute the maximum combined stress and the resulting design factor for the horizontal cantilever beams in Figure 10–15.

Given          Load weighs 40.0 kN. Force system detail shown in Figure 10–16. Tension in each cable is 40 000 N. The 75 cm long beams are steel IPE 180×91 made from A36 structural steel. From Appendix A–7(e), the properties of the beam are A = 2395 mm² and S = 1.463 × 10^{5}  mm^{3} .

Analysis    The tension in the cable attached to the end of each cantilever beam will tend to cause direct tensile stress combined with bending stress in the beam. Use the Guidelines for Solving Problems with Combined Normal Stresses outlined in this section.

Results       Step 1. Figure 10–17 shows the free-body diagram of one beam with the 40.0 kN force applied by the cable to the end of the beam. The reaction at the left end where the beam is rigidly attached to the column consists of a vertical reaction force, a horizontal reaction force, and a CCW moment.

Step 2. Also shown in Figure 10–17 is the resolution of the 40.0 kN force into vertical and horizontal components where F_{v} = 20 000 N and F_{h} = 34 600 N .

Step 3. The horizontal force, F_{h} , acts in a direction that is coincident with the neutral axis of the beam. Therefore, it causes direct tensile stress with a magnitude of

\sigma_{t} = \frac{F_{h}}{A} = \frac{34  600  N}{2395   mm^{2}}= 14.45 MPa

Step 4. The vertical force, F_{v} , causes bending in a downward direction such that the top of the beam is in tension and the bottom is in compression. The maximum bending moment will occur at the support at the left end, where

M = F_{v} (75 cm) = ( 20 000 N)(75 cm)(1 m / 100 cm) = 15 000 N·m

Then the maximum bending stress caused by this moment is

\sigma_{b} = \frac{M}{S} = \frac{15  000  N·m}{1.463 \times 10^{5}   mm^{3}}= 102.53 MPa

A stress of this magnitude occurs as a tensile stress at the top surface and as a compressive stress at the bottom surface of the beam at the support.

Step 5. This step does not apply to this problem because there is no horizontal force offset from the neutral axis.

Step 6. It can be reasoned that the maximum combined stress occurs at the top surface of the beam at the support because both the direct tensile stress computed in step 3 and the bending stress computed in step 4 are tensile at that point. Therefore, they will add together. Using superposition, as defined in Equation 10–1,

\sigma_{comb} = \pm \frac{F}{A} \pm \frac{M}{S}

For the combined stress at the top surface, called σ_{top} ,

σ_{top} = 14.45 MPa + 102.53 MPa = 116.98 MPa tension

For comparison, the combined stress at the bottom surface of the beam is

σ_{top} = 14.45 MPa – 102.53 MPa = – 88.08 MPa compression

Figure 10–18 shows a set of diagrams that illustrate the process of superposition. Part (a) is the stress in the beam caused by bending. Part (b) shows the direct tensile stress due toF_{h} . Part (c) shows the combined stress distribution.

Step 7. Because the load is static, we can compute the design factor based on the yield strength of the A36 steel, where s_{y} = 248 MPa (Appendix A–14). Then,

N = s_{y} / \sigma_{top} = (248 MPa)/(116.98 MPa) = 2.12

Comment   This should be adequate for a purely static load. If there is uncertainty about the magnitude of the load or if there is a possibility of the load being applied with some shock or impact, a higher design factor would be preferred.