Question 7.9: Figure 7–25 shows a cantilever bracket carrying a partial un...

Objective              Evaluate the beam shown in Figure 7–25 to ensure that the minimum design factor is 8 based on ultimate strength. If not, redesign the beam.

Given                    Loading and beam geometry in Figure 7–25; aluminum 7075-T6.

Analysis                1. The shearing force and bending moment diagrams will be drawn.

2. The design stress will be computed from σ_{d} = s_{u} /8.

3. The stress will be computed at sections A, B, and C, considering stress concentrations at B and C. These are the three likely points of failure because of either the magnitude of the bending moment or stress concentration. Everywhere else will have a lower bending stress. At each section, the stress will be computed from σ = MK_{t} /S, and the values of M, K_{t} , and S must be determined at each section.

4. The computed stresses will be compared with the design stress.

5. For any section having a stress higher than the design stress, a redesign will be proposed and the stress will be recomputed to verify that it is safe as redesigned.

Results                 Step 1. Figure 7–26 shows the completed shearing force and bending moment diagrams. Note that the values of the bending moment at sections B and C have been computed. You should verify the given values.

Step 2. From Appendix A–14, we find s_{u} = 572 MPa = 572 N/mm². Then,

\sigma_{d} = s_{u} /8 = 71.5 MPa

Steps 3 and 4. At each section, σ = MK_{t} /S = K_{t} σ_{nom}

Section A: K_{t} = 1.0 (given). M_{A} = 800 N · m

Dimensions b = 12.0 mm; h = 80.0 mm; rectangle

S_{A} = bh^{2} /6  = (12.0 mm)(80.0 mm)²/6 = 12 800 mm³

\sigma_{A} =\frac{(800  000  N·mm)(1.0)}{12  800  mm^{3}}  = 62.5 MPa < 71.5 MPa   OK

Section B: M_{B} = 403.1 N ∙ m. Find K_{t} the nominal stress, σ_{nom} from Appendix A–18–4. Note that the equation for , is given in the figure there, based on the net section modulus at the section considering the hole.

Dimensions t = b = 12 mm; w = h = 80 mm; d = 56 mm.

d/w = 56/80 = 0.7 = 0.692; K_{t} = 1.40 (curve C) .

\sigma_{B} = K_{t} \sigma_{nom} = \frac{K_{T}(6M_{B}w)}{(w^{3}-d^{3})t} = \frac{(1.40)(6)(403.1  N·m)(80  mm)\left\lgroup \frac{1000  mm}{1  m}\right\rgroup }{[(80  mm)^{3}-(56  mm)^{3}] (12  mm)}

= 67.1 MPa < 71. 5 MPa  OK

Section C: M_{C} = 162.5  N ∙ m. Find Kt from Appendix A–18–10.

Dimensions t = 12 mm; H = 80 mm; h = 50 mm; r = 2 mm.

H/h = 80/50 = 1.6; r/h = 2/50 =0.04;   then, K_{t} = 2.40.

Section modulus = S_{c} = th^{2} /6  = (12)(50)²/6 = 5000 mm³

\sigma_{c} = \frac{M_{c}K_{t}}{S_{c}} = \frac{(162.5  N·m)(2.40)\left\lgroup \frac{1000  mm}{1  m}\right\rgroup}{5000  mm³} = 78.0 MPa > 71.5 MPa  too high

Step 5. Proposed redesign at C: Because the stress concentration factor at section C is quite high, increase the fillet radius. The maximum allowable stress concentration factor is found by solving the stress equation for K_{t} and letting σ = σ_{d} = 71.5 MPa. Then,

K_{t} = \frac{S_{c}\sigma_{d}}{M} = \frac{(5000  mm³)(71.5  N/mm²)}{(162.5  N·m)\left\lgroup \frac{1000  mm}{1  m}\right\rgroup} = 2.20

From Appendix A–18–10, the minimum value of r/h = 0.05 to limit K_{t} to 2.20. Then,

r_{min} = 0.05(h) = 0.05(80) = 4

Let r = 4 mm; r/h = 4/50 = 0.08; K_{t} to 2.20. Then,

\sigma_{c} = \frac{M_{c}K_{t}}{S_{c}} = \frac{(162.5  N·m)\left\lgroup \frac{1000  mm}{1  m}\right\rgroup (2.20)}{5000  mm³} = 71.5 MPa > 71.5 MPa      OK

Comment           This problem is a good illustration of the necessity of analyzing any point within a beam where high stress may occur because of high bending moment, high stress concentration, small section modulus, or some combination of these. It also demonstrates one method of redesigning a beam to ensure safety.