A part of a conveyor system for a production operation is shown in Figure 5–17. The complete system will include several hundred hanger assemblies like this one. Design the horizontal bar that extends between two adjacent conveyor hangers and that supports a fixture at its midpoint.

Question

A part of a conveyor system for a production operation is shown in Figure 5–17. The complete system will include several hundred hanger assemblies like this one. Design the horizontal bar that extends between two adjacent conveyor hangers and that supports a fixture at its midpoint. The empty fixture weighs 85 lb. A cast iron engine block weighing 225 lb is hung on the fixture to carry it from one process to another, where it is then removed. It is expected that the bar will experience several thousand cycles of loading and unloading of the engine blocks. Design Example 5–2 considered this same system with the objective of specifying the diameter of the pins. The pin at the middle of the horizontal bar where the fixture is hung has been specified to have a diameter of 0.50 in. Those at each end where the horizontal bar is connected to the conveyor hangers are each 0.375 in.

Accepted Answer

Objective: Design the horizontal bar for the conveyor system.
Given: The general arrangement is shown in Figure 5–17. The bar is simply supported at points 24 in apart. A vertical load that is alternately 85 lb and 310 lb (85 + 225) is applied at the middle of the bar through the pin connecting the fixture to the bar. The load will cycle between these two values many thousands of times in the expected life of the bar. The pin at the middle of the bar has a diameter of 0.50 in, while the pins at each end are 0.375 in.

Basic Design Decisions:
It is proposed to make the bar from steel in the form of a rectangular bar with the long dimension of its cross section vertical. Cylindrical holes will be machined on the neutral axis of the bar at the support points and at its center to receive cylindrical pins that will attach the bar to the conveyor carriers and to the fixture. Figure 5–18 shows the basic design for the bar.
The thickness of the bar, [latex]t[/latex], should be fairly large to provide a good bearing surface for the pins and to ensure lateral stability of the bar when subjected to the bending stress. A relatively thin bar would tend to buckle along its top surface where the stress is compressive. As a design decision, we will use a thickness of [latex]t[/latex] = 0.50 in. The design analysis will determine the required height of the bar, [latex]h[/latex], assuming that the primary mode of failure is stress due to bending. Other possible modes of failure are discussed in the comments at the end of this example.
An inexpensive steel is desirable because several hundred bars will be made. We specify SAE 1020 hot-rolled steel having a yield strength of [latex]s_y[/latex] = 30 ksi and an ultimate strength of [latex]s_u[/latex] = 55 ksi (Appendix 3).

APPENDIX 3 Design Properties of Carbon and Alloy Steel

Material designation (SAE number)	Condition	Tensile strength		Yield strength		Ductility (percent elongation in 2 in)	Brinell hardness (HB)
Material designation (SAE number)	Condition	(ksi)	(MPa)	(ksi)	(MPa)	Ductility (percent elongation in 2 in)	Brinell hardness (HB)
1020	Hot-rolled	55	379	30	207	25	111
1020	Cold-drawn	61	420	51	352	15	122
1020	Annealed	60	414	43	296	38	121
1040	Hot-rolled	72	496	42	290	18	144
1040	Cold-drawn	80	552	71	490	12	160
1040	OQT 1300	88	607	61	421	33	183
1040	OQT 400	113	779	87	600	19	262
1050	Hot-rolled	90	620	49	338	15	180
1050	Cold-drawn	100	690	84	579	10	200
1050	OQT 1300	96	662	61	421	30	192
1050	OQT 400	143	986	110	758	10	321
1117	Hot-rolled	65	448	40	276	33	124
1117	Cold-drawn	80	552	65	448	20	138
1117	WQT 350	89	614	50	345	22	178
1137	Hot-rolled	88	607	48	331	15	176
1137	Cold-drawn	98	676	82	565	10	196
1137	OQT 1300	87	600	60	414	28	174
1137	OQT 400	157	1083	136	938	5	352
1144	Hot-rolled	94	648	51	352	15	188
1144	Cold-drawn	100	690	90	621	10	200
1144	OQT 1300	96	662	68	496	25	200
1144	OQT 400	127	876	91	627	16	277
1213	Hot-rolled	55	379	33	228	25	110
1213	Cold-drawn	75	517	58	340	10	150
12L13	Hot-rolled	57	393	34	234	22	114
12L13	Cold-drawn	70	483	60	414	10	140
1340	Annealed	102	703	63	434	26	207
1340	OQT 1300	100	690	75	517	25	235
1340	OQT 1000	144	993	132	910	17	363
1340	OQT 700	221	1520	197	1360	10	444
1340	OQT 400	285	1960	234	1610	8	578
3140	Annealed	95	655	67	462	25	187
3140	OQT 1300	115	792	94	648	23	233
3140	OQT 1000	152	1050	133	920	17	311
3140	OQT 700	220	1520	200	1380	13	461
3140	OQT 400	280	1930	248	1710	11	555
4130	Annealed	81	558	52	359	28	156
4130	WQT 1300	98	676	89	614	28	202
4130	WQT 1000	143	986	132	910	16	302
4130	WQT 700	208	1430	180	1240	13	415
4130	WQT 400	234	1610	197	1360	12	461
4140	Annealed	95	655	54	372	26	197
4140	OQT 1300	117	807	100	690	23	235
4140	OQT 1000	168	1160	152	1050	17	341
4140	OQT 700	231	1590	212	1460	13	461
4140	OQT 400	290	2000	251	1730	11	578
4150	Annealed	106	731	55	379	20	197
4150	OQT 1300	127	880	116	800	20	262
4150	OQT 1000	197	1360	181	1250	11	401
4150	OQT 700	247	1700	229	1580	10	495
4150	OQT 400	300	2070	248	1710	10	578
4340	Annealed	108	745	68	469	22	217
4340	OQT 1300	140	965	120	827	23	280
4340	OQT 1000	171	1180	158	1090	16	363
4340	OQT 700	230	1590	206	1420	12	461
4340	OQT 400	283	1950	228	1570	11	555
5140	Annealed	83	572	42	290	29	167
5140	OQT 1300	104	717	83	572	27	207
5140	OQT 1000	145	1000	130	896	18	302
5140	OQT 700	220	1520	200	1380	11	429
5140	OQT 400	276	1900	226	1560	7	534
5150	Annealed	98	676	52	359	22	197
5150	OQT 1300	116	800	102	700	22	241
5150	OQT 1000	160	1100	149	1030	15	321
5150	OQT 700	240	1650	220	1520	10	461
5150	OQT 400	312	2150	250	1720	8	601
5160	Annealed	105	724	40	276	17	197
5160	OQT 1300	115	793	100	690	23	229
5160	OQT 1000	170	1170	151	1040	14	341
5160	OQT 700	263	1810	237	1630	9	514
5160	OQT 400	322	2220	260	1790	4	627
6150	Annealed	96	662	59	407	23	197
6150	OQT 1300	118	814	107	738	21	241
6150	OQT 1000	183	1260	173	1190	12	375
6150	OQT 700	247	1700	223	1540	10	495
6150	OQT 400	315	2170	270	1860	7	601
8650	Annealed	104	717	56	386	22	212
8650	OQT 1300	122	841	113	779	21	255
8650	OQT 1000	176	1210	155	1070	14	363
8650	OQT 700	240	1650	222	1530	12	495
8650	OQT 400	282	1940	250	1720	11	555
8740	Annealed	100	690	60	414	22	201
8740	OQT 1300	119	820	100	690	25	241
8740	OQT 1000	175	1210	167	1150	15	363
8740	OQT 700	228	1570	212	1460	12	461
8740	OQT 400	290	2000	240	1650	10	578
9255	Annealed	113	780	71	490	22	229
9255	O&T 1300	130	896	102	703	21	262
9255	O&T 1000	181	1250	160	1100	14	352
9255	O&T 700	260	1790	240	1650	5	534
9255	O&T 400	310	2140	287	1980	2	601

Analysis: The Goodman criterion applies for completing the design analysis because fluctuating normal stress due to bending is experienced by the bar. Equation (5–32) ([latex]\frac{K_{t} \sigma_{a}^{\prime}}{s_{n}^{\prime}}+\frac{\sigma_{m}^{\prime}}{s_{u}}=\frac{1}{N}[/latex]) will be used:

[latex]\frac{1}{N}=\frac{\sigma_{m}}{s_{u}}+\frac{K_{t} \sigma_{a}}{s_{n}^{\prime}}[/latex]

In general, the bending stress in the bar will be computed from the flexure formula:
[latex]\sigma=M / S[/latex]
where [latex]M[/latex] = bending moment
[latex]S[/latex] = section modulus of the cross section of the bar = [latex]th^2[/latex]/6 (Appendix 1)

APPENDIX 1 Properties of Areas
[latex]\begin{array}{ll}A= ext { area } & r= ext { radius of gyration }=\sqrt{I / A} \I= ext {moment of inertia } & J= ext { polar moment of inertia } \S= ext { section modulus } &Z_{p}= ext { polar section modulus }\end{array}[/latex]
	[latex]\begin{array}{ll}A=\pi D^{2} / 4 & r=D / 4 \I=\pi D^{4} / 64 & J=\pi D^{4} / 32 \S=\pi D^{3} / 32 & Z_{p}=\pi D^{3} / 16\end{array}[/latex]
	[latex]\begin{array}{ll}A=\pi\left(D^{2}-d^{2} ight) / 4 & r=\frac{\sqrt{\left(D^{2}+d^{2} ight)}}{4} \I=\pi\left(D^{4}-d^{4} ight) / 64 & J=\pi\left(D^{4}-d^{4} ight) / 32 \S=\pi\left(D^{4}-d^{4} ight) / 32 D & Z_{p}=\pi\left(D^{4}-d^{4} ight) / 16 D\end{array}[/latex]
	[latex]\begin{aligned}&A=H^{2} \quad\quad\quad r=H / \sqrt{12} \&I=H^{4} / 12 \&S=H^{3} / 6\end{aligned}[/latex]
	[latex]\begin{array}{ll}A=B H & r_{x}=H / \sqrt{12} \I_{x}=B H^{3} / 12 & r_{y}=B / \sqrt{12} \I_{y}=H B^{3} / 12 & \S_{x}=B H^{2} / 6 & \S_{y}=H B^{2} / 6 &\end{array}[/latex]
	[latex]\begin{aligned}&A=B H-b h \&I_{x}=\frac{B H^{3}-b h^{3}}{12} \quad S_{x}=\frac{B H^{3}-b h^{3}}{6 H}\quad r_{x}=0.289 \sqrt{\frac{B H^{3}-b h^{3}}{B H-b h}} \&I_{y}=\frac{H B^{3}-h b^{3}}{12} \quad S_{y}=\frac{H B^{3}-h b^{3}}{6 B} \quad r_{y}=0.289 \sqrt{\frac{H B^{3}-h b^{3}}{H B-h b}}\end{aligned}[/latex]
	[latex]\begin{aligned}&A=B H / 2 \quad\quad\quad r=H / \sqrt{18} \&I=B H^{3} / 36 \&S=B H^{2} /24\end{aligned}[/latex]
	[latex]\begin{aligned}&A=\pi D^{2} / 8 \quad\quad\quad r=0.132 D \&I=0.007 D^{4} \&S=0.024D^{3}\end{aligned}[/latex]
	[latex]\begin{aligned}&A=0.866 D^{2} \quad\quad\quad r=0.264 D \&I=0.06 D^{4} \&S=0.12 D^{3}\end{aligned}[/latex]
	[latex]\begin{aligned}A &=H(a+B) / 2 \y &=\frac{H(a+2 B)}{3(a+B)} \quad\quad\quad\quad\quad\quad S=\frac{H^{2}\left(a^{2}+4 a B+B^{2} ight)}{12(a+2 B)} \I_{x} &=\frac{H^{3}\left(a^{2}+4a B+B^{2} ight)}{36(a+B)} \quad\quad r=\frac{H^{2}\left(a^{2}+4 a B+B^{2} ight)}{18(a+B)^{2}} \y &=\underset{ ext { Maximum distance from } x ext {-axis to }}{ ext { outer surface of section }}\end{aligned}[/latex]
	[latex]\begin{aligned}&A=\pi b h \&I=\frac{\pi h^{3} b}{4} \&S=\frac{\pi h^{2} b}{4} \&r=h / 2\end{aligned}[/latex]

Our approach will be to first determine the values for both the mean and the alternating bending moments experienced by the bar at its middle. Then the yield and endurance limit values for the steel will be found. Reference 5 in Chapter 3 indicates that a small hole, with diameter d, in a plate-beam does not weaken the beam if the ratio [latex]d/h[/latex] is less than 0.50. That is, if [latex]d/h[/latex] 6 0.50, [latex]K_t[/latex] = 1.0. We will make that assumption and check it later. Based on the application conditions, let’s use [latex]N[/latex] = 4 as advised in item 4 in Section 5–9 because the actual use pattern for this conveyor system in a factory environment is somewhat uncertain and shock loading is likely.

Bending Moments Figure 5–18 shows the shearing force and bending moment diagrams for the bar when carrying just the fixture and then both the fixture and the engine block. The maximum bending moment occurs at the middle of the bar where the load is applied. The values are [latex]M_{max}[/latex] = 1860 lb . in with the engine block on the fixture and [latex]M_{min}[/latex] = 510 lb . in for the fixture alone. Now the values for the mean and alternating bending moments are calculated using modified forms of Equations (5–1) ([latex]\sigma_{m}=\left(\sigma_{\max }+\sigma_{\min } ight) / 2[/latex]) and (5–2) ([latex]\sigma_{a}=\left(\sigma_{\max }-\sigma_{\min } ight) / 2[/latex]):

[latex]\begin{aligned}&M_{m}=\left(M_{\max }+M_{\min } ight) / 2=(1860+510) / 2=1185\mathrm{lb} \cdot \mathrm{in} \&M_{a}=\left(M_{\max }-M_{\min } ight) /2=(1860-510) / 2=675 \mathrm{lb} \cdot \mathrm{in}\end{aligned}[/latex]
The stresses will be found from [latex]\sigma_{m}=\frac{M_{m}}{S}[/latex] and [latex]\sigma_{a}=\frac{M_{a}}{S}[/latex]. Note that the stress element at the location of interest is subjected to uniaxial tension/compression without any shear. For [latex]1 \mathrm{D}[/latex] loading.
[latex]\begin{aligned}&\sigma_{m 1}=\sigma_{m}=\frac{M_{m}}{S}, \sigma_{m 2}=\sigma_{m 3}=0 \&\sigma_{a 1}=\sigma_{a}=\frac{M_{a}}{S}, \sigma_{a 2}=\sigma_{a 3}=0\end{aligned}[/latex]
As such, we can now apply the Goodman criterion from Equation (5-32) ([latex]\frac{K_{t} \sigma_{a}^{\prime}}{s_{n}^{\prime}}+\frac{\sigma_{m}^{\prime}}{s_{u}}=\frac{1}{N}[/latex]) with
[latex]\begin{gathered}\sigma_{m}^{\prime}=\sigma_{m}=\frac{M_{m}}{S} \\sigma_{a}^{\prime}=\sigma_{a}=\frac{M_{a}}{S}\end{gathered}[/latex]

Material Strength Values The material strength properties required are the ultimate strength [latex]s_{u}[/latex] and the, estimated actual endurance limit [latex]s_{n}^{\prime}[/latex]. We know that the ultimate strength [latex]s_{u}=55 \mathrm{ksi}[/latex]. We now find [latex]s_{n}^{\prime}[/latex] using the method outlined in Section 5-6.

Size factor, [latex]C s:[/latex] From Section 5-6, Equation (5-10) ([latex]\sigma_{e} \leq \frac{s_{y}}{N}[/latex]) defines an equivalent diameter, [latex]D_{e}[/latex], for the rectangular section as
[latex]D_{e}=0.808 \sqrt{h t}[/latex]
We have specified the thickness of the bar to be [latex]t=0.50[/latex] in. The height is unknown at this time. As an estimate, let's assume [latex]h \approx 2.0 \mathrm{in}[/latex]. Then,
[latex]D_{e}=0.808 \sqrt{h t}=0.808 \sqrt{(2.0 \mathrm{in})(0.50 \mathrm{in})}=0.808 \mathrm{in}[/latex]
We can now use Figure 5-12 or the equations in Table 5-4 to find [latex]C_{s}=0.90[/latex]. This value should be checked later after a specific height dimension is proposed.

TABLE 5–4 Size Factors
U.S. customary units
Size rangeFor D in inches
[latex]D \leq 0.30[/latex]	[latex]C_{s}=1.0[/latex]
[latex]0.30	[latex]C_{s}=(D / 0.3)^{-0.11}[/latex]
[latex]2.0	[latex]C_{s}=0.859-0.02125 D[/latex]
SI units
Size range	For D in mm
[latex]D \leq 7.62[/latex]	[latex]C_{s}=1.0[/latex]
[latex]7.62	[latex]C_{s}=(D / 7.62)^{-0.11}[/latex]
[latex]50	[latex]C_{s}=0.859-0.000837 D[/latex]

Material factor,[latex]C_{m}:[/latex] Use [latex]C_{m}=1.0[/latex] for the wrought, hot-rolled steel.
Stress-type factor, [latex]C_{ ext {sf }}[/latex] Use [latex]C_{ ext {st }}=1.0[/latex] for repeated bending stress.
Reliability factor, [latex]C_{R}:[/latex] A high reliability is desired. Let's use [latex]C_{R}=0.75[/latex] to achieve a reliability of 0.999 as indicated in Table 5-3.

Table 5–3 Approximate Reliability Factors, [latex]C_R[/latex]
Desired reliability	[latex]C_R[/latex]
0.5	1
0.9	0.91
0.99	0.81
0.999	0.75

The value of [latex]s_n[/latex] = 20 ksi is found from Figure 5–11 for hot-rolled steel having an ultimate strength of 55 ksi.
Now, applying Equation (5–21) ([latex]s_{n}^{\prime}=s_{n}\left(C_{m} ight)\left(C_{s t} ight)\left(C_{R} ight)\left(C_{s} ight)[/latex]) from Section 5–6, we have

[latex]s_{n}^{\prime}=\left(C_{m} ight)\left(C_{s t} ight)\left(C_{R} ight)\left(C_{s} ight) s_{n}=(1.0)(1.0)(0.75)(0.90)(20 \mathrm{ksi})=13.5 \mathrm{ksi}[/latex]

Solution for the Required Section Modulus At this point, we have specified all factors in Equation (5–32) ([latex]\frac{K_{t} \sigma_{a}^{\prime}}{s_{n}^{\prime}}+\frac{\sigma_{m}^{\prime}}{s_{u}}=\frac{1}{N}[/latex]) except the section modulus of the cross section of the bar that is involved in each expression for stress
as shown above. We will now solve the equation for the required value of [latex]S[/latex].
Recall that we showed earlier that [latex]\sigma_{m}=M_{m} / S ext { and } \sigma_{a}=M_{d} / S ext {. }[/latex] Then

[latex]\begin{aligned}\frac{1}{N} &=\frac{\sigma_{m}}{s_{u}}+\frac{K_{t} \sigma_{a}}{s_{n}^{\prime}}=\frac{M_{m}}{S s_{u}}+\frac{K_{t} M_{a}}{S s_{n}^{\prime}}=\frac{1}{S}\left[\frac{M_{m}}{s_{u}}+\frac{K_{t} M_{a}}{s_{n}^{\prime}} ight] \S&=N\left[\frac{M_{m}}{s_{u}}+\frac{K_{t} M_{a}}{s_{n}^{\prime}} ight]=4\left[\frac{1185 \mathrm{lb} \cdot \mathrm{in}}{55000 \mathrm{lb} / \mathrm{in}^{2}}+\frac{1.0(675 \mathrm{lb} \cdot \mathrm{in})}{13500 \mathrm{lb} / \mathrm{in}^{2}} ight] \S &=0.286 \mathrm{in}^{3}\end{aligned}[/latex]

Results: The required section modulus has been found to be [latex]S = 0.286 in^3[/latex]. We observed earlier that [latex]S = th^2/6[/latex] for a solid rectangular cross section, and we decided to use this form to find an initial estimate for the required height of the section, [latex]h[/latex]. We have specified [latex]t[/latex] = 0.50 in. Then the estimated minimum acceptable value for the height [latex]h[/latex] is

[latex]h=\sqrt{6 S / t}=\sqrt{6\left(0.286 \mathrm{in}^{3} ight) /(0.50 \mathrm{in})}=1.85 \mathrm{in}[/latex]

The table of preferred basic sizes in the decimal-inch system (Table A2–1) recommends [latex]h[/latex] = 2.00 in.

TABLE A2–1 Preferred Basic Sizes
Fractional (in)		Decimal (in)	SI metric (mm)
[latex]\begin{array}{ll}1 / 64 & 0.015625 \1 / 32 & 0.03125 \1 / 16 & 0.0625 \3 / 32 & 0.09375\1 / 8 & 0.1250\end{array}[/latex]	[latex]\begin{array}{ll}5 & 5.000 \5 \frac{1}{4} & 5.250 \5 \frac{1}{2} & 5.500 \5 \frac{3}{4}& 5.750 \6 & 6.000\end{array}[/latex]	[latex]\begin{array}{rrr}0.010 & 2.00 & 8.50 \0.012 & 2.20 & 9.00 \0.016 & 2.40 & 9.50 \0.020&2.60 & 10.00 \0.025 & 2.80 & 10.50\end{array}[/latex]	[latex]\begin{array}{ll}1.0 & 40 \1.1 & 45 \1.2 & 50 \1.4 & 55 \1.6 & 60\end{array}[/latex]
[latex]\begin{array}{cc}5 / 32 & 0.15625 \3 / 16 & 0.1875 \1 / 4 & 0.2500 \5 / 16 & 0.3125 \3/ 8 & 0.3750\end{array}[/latex]	[latex]\begin{array}{ll}6 \frac{1}{2} & 6.500 \7 & 7.000 \7 \frac{1}{2} & 7.500 \8 & 8.000 \8\frac{1}{2} & 8.500\end{array}[/latex]	[latex]\begin{array}{lll}0.032 & 3.00 & 11.00 \0.040 & 3.20 & 11.50 \0.05 & 3.40 & 12.00 \0.06& 3.60 & 12.50 \0.08 & 3.80 & 13.00\end{array}[/latex]	[latex]\begin{array}{ll}1.8 & 70 \2.0 & 80 \2.2 & 90 \2.5 & 100 \2.8 & 110\end{array}[/latex]
[latex]\begin{array}{cc}7 / 16 & 0.4375 \1 / 2 & 0.5000 \9 / 16 & 0.5625 \5 / 8 & 0.6250\end{array}[/latex]	[latex]\begin{array}{lr}9 & 9.000 \9 \frac{1}{2} & 9.500 \10 & 10.000 \10 \frac{1}{2} & 10.500\end{array}[/latex]	[latex]\begin{array}{lll}0.10 & 4.00 & 13.50 \0.12 & 4.20 & 14.00 \0.16 & 4.40 & 14.50 \0.20 &4.60 & 15.00\end{array}[/latex]	[latex]\begin{array}{ll}3.0 & 120 \3.5 & 140 \4.0 & 160 \4.5 & 180\end{array}[/latex]
[latex]\begin{array}{cc}11 / 16 & 0.6875 \3 / 4 & 0.7500 \7 / 8 & 0.8750\end{array}[/latex]	[latex]\begin{array}{ll}11 & 11.000 \11 \frac{1}{2} & 11.500 \12 & 12.000\end{array}[/latex]	[latex]\begin{array}{lll}0.24 & 4.80 & 15.50 \0.30 & 5.00 & 16.00 \0.40 & 5.20 & 16.50\end{array}[/latex]	[latex]\begin{array}{ll}5.0 & 200 \5.5 & 220 \6 & 250\end{array}[/latex]
[latex]\begin{array}{cc}1 & 1.000 \1 \frac{1}{4} & 1.250 \1 \frac{1}{2} & 1.500 \1 \frac{3}{4}& 1.750\end{array}[/latex]	[latex]\begin{array}{ll}12 \frac{1}{2} & 12.500 \13 & 13.000 \13 \frac{1}{2} & 13.500 \14 &14.000\end{array}[/latex]	[latex]\begin{array}{lll}0.50 & 5.40 & 17.00 \0.60 & 5.60 & 17.50 \0.80 & 5.80 & 18.00 \1.00 &6.00 & 18.50\end{array}[/latex]	[latex]\begin{array}{rr}7 & 280 \8 & 300 \9 & 350 \10 & 400\end{array}[/latex]
[latex]\begin{array}{cc}2 & 2.000 \2 \frac{1}{4} & 2.250 \2 \frac{1}{2} & 2.500 \2 \frac{3}{4}& 2.750\end{array}[/latex]	[latex]\begin{array}{ll}14 \frac{1}{2} & 14.500 \15 & 15.000 \15 \frac{1}{2} & 15.500 \16 &16.000\end{array}[/latex]	[latex]\begin{array}{lll}1.20 & 6.50 & 19.00 \1.40 & 7.00 & 19.50 \1.60 & 7.50 & 20.00 \1.80 &8.00 &\end{array}[/latex]	[latex]\begin{array}{ll}11 & 450 \12 & 500 \14 & 550 \16 & 600\end{array}[/latex]
[latex]\begin{array}{ll}3 & 3.000 \3 \frac{1}{4} & 3.250 \3 \frac{1}{2} & 3.500 \3 \frac{3}{4}& 3.750\end{array}[/latex]	[latex]\begin{array}{ll}16 \frac{1}{2} & 16.500 \17 & 17.000 \17 \frac{1}{2} & 17.500 \18 &18.000\end{array}[/latex]		[latex]\begin{array}{cc}18 & 700 \20 & 800 \22 & 900 \25 & 1000\end{array}[/latex]
[latex]\begin{array}{cc}4 & 4.000 \4 \frac{1}{4} & 4.250 \4 \frac{1}{2} & 4.500 \4 \frac{3}{4}& 4.750\end{array}[/latex]	[latex]\begin{array}{ll}18 \frac{1}{2} & 18.500 \19 & 19.000 \19 \frac{1}{2} & 19.500 \20 &20.000\end{array}[/latex]		[latex]\begin{aligned}&28 \&30 \&35\end{aligned}[/latex]

We should first check the earlier assumption that the ratio [latex]d/h < 0.50[/latex] at the middle of the bar. The actual ratio is

[latex]d / h=(0.50 \mathrm{in}) /(2.00 \mathrm{in})=0.25 ext { (okay) }[/latex]
This indicates that our earlier assumption that [latex]K_{t}=1.0[/latex] is correct. Also, our assumed value of [latex]C_{s}=0.90[/latex] is correct because the actual height, [latex]h=2.0[/latex] in, is identical to our assumed value.

We will now compute the actual value for the section modulus of the cross section with the hole in it. See Figure A15-6 in the appendix.
[latex]S=\frac{t\left(h^{3}-d^{3} ight)}{6 h}=\frac{(0.50 \mathrm{in})\left[(2.00 \mathrm{in})^{3}-(0.50 \mathrm{in})^{3} ight]}{6(2.00 \mathrm{in})}=0.328 \mathrm{in}^{3}[/latex]
This value is larger than the minimum required value of [latex]0.286 \mathrm{in}^{3}[/latex]. Therefore, the size of the cross section is satisfactory with regard to stress due to bending.

Final Design Decisions and Comments

In summary, the following are the design decisions for the horizontal bar of the conveyor hanger shown in Figure 5–18.

1. Material: SAE 1020 hot-rolled steel.
2. Size: Rectangular cross section. Thickness [latex]t[/latex] = 0.50 in; height [latex]h[/latex] = 2.00 in.
3. Overall design: Figure 5–18 shows the basic features of the bar.
4. Other considerations: Remaining to be specified are the tolerances on the dimensions for the bar and the finishing of its surfaces. The potential for corrosion should be considered and may call for paint, plating, or some other corrosion protection. The size of the cross section can likely be
used with the as-received tolerances on thickness and height, but this is somewhat dependent on the design of the fixture that holds the engine block and the conveyor hangers. So the final tolerances will be left open pending later design decisions. The holes in the bar for the pins should be designed to produce a close sliding fit with the pins, and the details of specifying the tolerances on the hole diameters for such a fit are discussed in Chapter 13. See also the discussion on lug joints in Section 3–21.
5. Other possible modes of failure: The analysis used in this problem assumed that failure would occur due to the bending stresses in the rectangular bar. The dimensions were specified to preclude this from happening. Other possible modes are discussed as follows:
a. Deflection of the bar as an indication of stiffness: The type of conveyor system described in this problem should not be expected to have extreme rigidity because moderate deflection of members should not impair its operation. However, if the horizontal bar deflects so much that it appears to be rather flexible, it would be deemed unsuitable. This is a subjective judgment. We can use Case (a) in Table A14–1 to compute the deflection.

TABLE A14–1 Beam-Deflection Formulas for Simply Supported Beams
[latex]\mathrm{y}_{B}=\mathrm{y}_{\max }=\frac{-P L^{3}}{48 E I} \quad ext { at center }[/latex] Between [latex]A[/latex] and [latex]B[/latex]; [latex]\mathrm{y}=\frac{-P x}{48 E l}\left(3 L^{2}-4 x^{2} ight)[/latex]
[latex]\begin{gathered}y_{\max }=\frac{-P a b(L+b) \sqrt{3 a(L+b)}}{27 E I L} \ ext { at }x_{1}=\sqrt{a(L+b) / 3} \y_{B}=\frac{-P a^{2} b^{2}}{3 E / L} ext { at load }\end{gathered}[/latex] Between[latex]A[/latex] and [latex]B[/latex] (the longer segment): [latex]y=\frac{-P b x}{6 E / L}\left(L^{2}-b^{2}-x^{2} ight)[/latex] Between [latex]B[/latex] and [latex]C[/latex] (the shorter segment): [latex]y=\frac{-P a v}{6 E I L}\left(L^{2}-v^{2}-a^{2} ight)[/latex] At end of overhang at [latex]D[/latex] : [latex]y_{D}=\frac{P a b c}{6 E I L}(L+a)[/latex]
[latex]\begin{gathered}y_{E}=y_{\max }=\frac{-P a}{24 E I}\left(3 L^{2}-4 a^{2} ight) ext {at center } \y_{B}=y_{C}=\frac{-P a^{2}}{6 E I}(3 L-4 a) ext { at loads }\end{gathered}[/latex] Between [latex]A[/latex] and [latex]B[/latex] : [latex]y=\frac{-P x}{6 E I}\left(3 a L-3 a^{2}-x^{2} ight)[/latex] Between [latex]B[/latex] and [latex]C[/latex] : [latex]y=\frac{-P a}{6 E I}\left(3 L x-3 x^{2}-a^{2} ight)[/latex]
[latex]y_{B}=y_{\max }=\frac{-5 w L^{4}}{384 E l}=\frac{-5 W L^{3}}{384 E I} ext { at center }[/latex] Between [latex]A[/latex] and [latex]B[/latex] : [latex]y=\frac{-w x}{24 E l}\left(L^{3}-2 L x^{2}+x^{3} ight)[/latex] At [latex]D[/latex] at end: [latex]y_{D}=\frac{w L^{3} a}{24 E I}[/latex]
Between [latex]A[/latex] and [latex]B[/latex] : [latex]y=\frac{-w x}{24 E I L}\left[a^{2}(2 L-a)^{2}-2 a x^{2}(2 L-a)+L x^{3} ight][/latex] Between [latex]B[/latex] and [latex]C[/latex] : [latex]y=\frac{-w a^{2}(L-x)}{24 E I L}\left(4 L x-2 x^{2}-a^{2} ight)[/latex]
[latex]M_{B}= ext { concentrated moment at } B[/latex] Between [latex]A[/latex] and [latex]B[/latex] : [latex]y=\frac{-M_{B}}{6 E I}\left[\left(6 a-\frac{3 a^{2}}{L}-2 L ight) x-\frac{x^{3}}{L} ight][/latex] Between [latex]B[/latex] and [latex]C[/latex] : [latex]y=\frac{M_{B}}{6 E I}\left[3 a^{2}+3 x^{2}-\frac{x^{3}}{L}-\left(2 L+\frac{3 a^{2}}{L} ight) x ight][/latex]
At [latex]C[/latex] at end of overhang: [latex]y_{C}=\frac{-P a^{2}}{3 E l}(L+a)[/latex] At [latex]D[/latex], maximum upward deflection: [latex]y_{D}=0.06415 \frac{P a L^{2}}{E l}[/latex]
At [latex]C[/latex] at center: [latex]y=\frac{-W L-2 a)^{3}}{384 E l}\left[\frac{5}{L}(L-2 a)-\frac{24}{L}\left(\frac{a^{2}}{L-2 a} ight) ight][/latex] At [latex]A[/latex] and [latex]E[/latex] at ends: [latex]y=\frac{-W L-2 a)^{3} a}{24 E I L}\left[-1+6\left(\frac{a}{L-2 a} ight)^{2}+3\left(\frac{a}{L-2 a} ight)^{3} ight][/latex]
At [latex]C[/latex] at center: [latex]y=\frac{P L^{2} a}{8 E I}[/latex] At [latex]A[/latex] and [latex]E[/latex] at ends at loads: [latex]y=\frac{-P a^{2}}{3 E I}\left(a+\frac{3}{2} L ight)[/latex]
At [latex]B[/latex] : [latex]y=0.03208 \frac{w a^{2} L^{2}}{E l}[/latex] At [latex]D[/latex] at end: [latex]y=\frac{-w a^{3}}{24 E l}(4 L+3 a)[/latex]

[latex]y=F L^{3} / 48 E I[/latex]
In this design,
[latex]\begin{aligned}&F=310 \mathrm{lb}= ext { maximum load on the bar } \&L=24.0 \mathrm{in}= ext { distance between supports } \&E=30 imes 10^{6} \mathrm{psi}= ext { modulusof elasticity of steel } \&I=t h^{3} / 12= ext { moment of inertia of the cross section }\&I=(0.50 \mathrm{in})(2.00 \mathrm{in})^{3} / 12=0.333 \mathrm{in}^{4}\end{aligned}[/latex]
Then,
[latex]y=\frac{(310 \mathrm{lb})(24.0 \mathrm{in})^{3}}{48\left(30 imes 10^{6} \mathrm{lb} / \mathrm{in}^{2} ight)\left(0.333 \mathrm{in}^{4} ight)}=0.0089 \mathrm{in}[/latex]
This value seems satisfactory. In Section 5-11, some guidelines were given for deflection of machine elements. One stated that bending deflections for general machine parts should be limited to the range of 0.0005 to 0.003 in/in of beam length. For the bar in this design, the ratio of [latex]y /L[/latex] can be compared to this range:
[latex]y /L[/latex] = (0.0089 in)/(24.0 in) = 0.0004 in/in of beam length

Therefore, this deflection is well within the recommended range.
b. Buckling of the bar: When a beam with a tall, thin, rectangular cross section is subjected to bending, it would be possible for the shape to distort due to buckling before the bending stresses would cause failure of the material. This is called elastic instability, and a complete discussion is beyond the scope of this book. However, Reference 14 shows a method of computing the critical buckling load for this kind of loading. The pertinent geometrical feature is the ratio of the thickness [latex]t[/latex] of the bar to its height [latex]h[/latex]. It can be shown that the bar as designed will not buckle.
c. Bearing stress on the inside surfaces of the holes in the beam: Pins transfer loads between the bar and the mating elements in the conveyor system. It is possible that the bearing stress at the pin–hole interface could be large, leading to excessive deformation or wear. Reference 4 in Chapter 3 indicates that the allowable bearing stress for a steel pin in a steel hole is 0.90[latex]s_y[/latex].

[latex]\sigma_{b d}=0.90 \mathrm{~s}_{y}=0.90(30000 \mathrm{psi})=27000 \mathrm{psi}[/latex]
The actual bearing stress at the center hole is found using the projected area, [latex]D_{p} t[/latex].
[latex]\sigma_{b}=F / D_{p} t=(310 \mathrm{lb}) /(0.50 \mathrm{in})(0.50 \mathrm{in})=1240 \mathrm{psi}[/latex]
Thus the pin and hole are very safe for bearing.

Question 5.DE.3: A part of a conveyor system for a production operation is sh...

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