Question 5.DE.4: A bracket is made by welding a rectangular bar to a circular...

Objective      The design process will be divided into two parts:

1. Design the rectangular bar for the bracket.

2. Design the circular rod for the bracket.

Rectangular Bar

Given     The bracket design is shown in Figure 5_20.The rectangular bar carries a load of 250lb vertically downward at its end. Support is provided by the weld at its left end where the loads are transferred to the circular rod. The bar acts as a cantilever beam, 12 in long. The design task is to specify the material for the bar and the dimensions of its cross section.

Basic Design Decisions      We will use steel for both parts of the bracket because of its relatively high stiffness, the ease of welding, and the wide range of strengths available. Let’s specify AISI 1340 annealed steel havings_{y}= 63 ksi ands_{u}= 102 ksi (Appendix 3). The steel is highly ductile, with a 26%f elongation.The objective of the design analysis that follows is to determine the size of the cross section of the rectangular bar. Assuming that the loading and processing conditions are well known,we will use a design factor of N=2 because of the static load.

Analysis and Results        The free-body diagram of the cantilever bar is shown in Figure 5_21. along with the shearing force and bending moment diagrams. This should be a familiar case, leading to the judgment that the maximum tensile stress occurs at the top of the bar near to where it is supported by the circular rod. This point is labeled element A in Figure 5_21. The maximum bending moment there is M = 3000 lb· in. The stress at A is \sigma _{A}=M/S where S = section modulus of the cross section of the bar.We will first compute the minimum allowable value for S and then determine the dimensions for the cross section. Case Cl from Section 5_9 applies because of the static loading. We will first compute the design stress from \sigma _{d}=s_{y} /N

\sigma_{d}=s_{y} / N=(63000 \mathrm{psi}) / 2=31500 \mathrm{psi} Now we must ensure that the expected maximum stress \sigma _{A}=M/S does not exceed the design stress. We can substitute \sigma _{A}=\sigma _{d} and solve for S. S=M / \sigma_{d}=(3000 \mathrm{lb} \cdot \mathrm{in}) /\left(31.500 \mathrm{lb} / \mathrm{in}^{2}\right)=0.095 \mathrm{in}^{3} The relationship for S is S=th^{2}/6As a design decision, let’s specify the approximate proportion for the cross-sectional dimensions
to be h= 3t. Then S=t h^{2} / 6=t(3 t)^{2} / 6=9 t^{3} / 6=1.5 t^{3} The required minimum thickness is then t=\sqrt[3]{S / 1.5}=\sqrt[3]{\left(0.095 \mathrm{in}^{3}\right) / 1.5}=0.399 \mathrm{in} The nominal height of the cross section should be, approximately,h=3t=3\left(0.399in\right)=1.20in

Final Design Decisions and Comments           In the fractional-inch system, standard sizes are selected to be t = 3/8 in = 0.375 in and h=\mid \frac{1}{4}in=1.25in (see Table A2_1). Note that we chose a slightly smaller value for t but a slightly larger value for h. We must check to see that the resulting value for S is satisfactory. S=t h^{2} / 6=(0.375 \mathrm{in})(1.25 \mathrm{in})^{2} / 6=0.0977 \mathrm{in}^{3} This is larger than the required value of 0.095in^{3} .so the design is satisfactory.

Circular Rod

Given        The bracket design is shown in Figure 5_20. The design task is to specify the material for the rod and the diameter of its cross section.

Basic Design Decisions          Let’s specify AISI 1340 annealed steel, the same as that used for the rectangular bar. Its properties are s_{y} = 63 ksi and s_{u} = 102 ksi.

Analysis and Results          Figure 5_22 is the free-body diagram for the rod. The rod is loaded al its left end by the reactions at the end of the rectangular bar, namely, a downward force of 250 lb and a moment of 3000 lb · in. The figure shows that the moment acts as a torque on the circular rod, and the 250-lb force causes bending with a maximum bending moment of 2000 Ib·in at the right end. Reactions are provided by the weld at its right end where the loads are transferred to the support. The rod then is subjected to a combined stress due to torsion and bending. Element B on the top of the rod is subjected to the maximum combined stress. The manner of loading on the circular rod is identical to that analyzed earlier in Section 4_6 in Chapter 4. It was shown that when only bending and torsional shear occur, a procedure called the equivalent torque method can be used to complete the analysis. First we define the equivalent torque,T_{e}:T_{e}=\sqrt{M^{2}+T^{2}}=\sqrt{(2000)^{2}+(3000)^{2}}=3606 \mathrm{lb} \cdot \mathrm{in} Then the shear stress in the bar is \tau =T_{e}/Z_{p} whereZ_{p}= polar section modulus For a solid circular rod,Z_{p}=\pi D^{3}/16 Our approach is to determine the design shear stress and T_{e} and then solve forZ_{p}, Case C2 using the maximum shear stress theory of failure can be applied. The design shear stress is \tau_{d}=0.50 s_{y} / N=(0.5)(63000 \mathrm{psi}) / 2=15750 \mathrm{psi} We let τ= T_{d} and solve forZ_{p}: Z_{p}=T_{e} / \tau_{d}=(3606 \mathrm{lb} \cdot \mathrm{in}) /\left(15750 \mathrm{lb} / \mathrm{in}^{2}\right)=0.229 \mathrm{in}^{3} Now that we know Z_{p}, we can compute the required diameter from  D={ }^{3} \sqrt{16 Z_{p} / \pi}=\sqrt[3]{16\left(0.229 \mathrm{in}^{3}\right) / \pi}=1.053 \mathrm{in} This is the minimum acceptable diameter for the rod.

Final Design Decisions and Comments        The circular rod is to be welded to the side of the rectangular bar,and we have specified the height of the bar to be\mid \frac{1}{4}in. Let’s specify the diameter of the circular rod to be machined to 1.10 in. This will allow welding all around its periphery.