For the flat steel spring shown in Figure 5–6, compute the maximum stress, the minimum stress, the mean stress, and the alternating stress. Also compute the stress ratio, R. The length L is 65 mm. The dimensions of the spring cross section are t = 0.80 mm and b = 6.0 mm.

Question

For the flat steel spring shown in Figure 5–6, compute the maximum stress, the minimum stress, the mean stress, and the alternating stress. Also compute the stress ratio, [latex]R[/latex]. The length [latex]L[/latex] is 65 mm. The dimensions of the spring cross section are [latex]t[/latex] = 0.80 mm and b = 6.0 mm.

Accepted Answer

Objective: Compute the maximum, minimum, mean, and alternating tensile stresses in the flat spring. Compute the stress ratio, [latex]R[/latex].

Given: Layout shown in Figure 5–6. The spring is steel: [latex]L[/latex] = 65 mm.
Spring cross-sectional dimensions: [latex]t[/latex] = 0.80 mm and b = 6.0 mm.
Maximum deflection of the spring at the follower = 8.0 mm.
Minimum deflection of the spring at the follower = 3.0 mm.

Analysis: Point [latex]A[/latex] at the base of the spring experiences the maximum tensile stress. Determine the force exerted on the spring by the follower for each level of deflection using the formulas from Table A14–2, Case (a).
Compute the bending moment at the base of the spring for each deflection. Then compute the stresses at point [latex]A[/latex] using the bending stress formula, [latex]\sigma=M c / I[/latex]. Use Equations (5–1) ([latex]\sigma_{m}=\left(\sigma_{\max }+\sigma_{\min } ight) / 2[/latex]), to (5–3) ([latex] ext { Stress ratio } R=\frac{ ext { minimum stress }}{ ext { maximum stress }}=\frac{\sigma_{\min }}{\sigma_{\max }}[/latex], [latex] ext { Stress ratio } A=\frac{ ext { alternating stress }}{ ext { mean stress }}=\frac{\sigma_{a}}{\sigma_{m}}[/latex]) for computing the mean, alternating stresses, and [latex]R[/latex].

Results: Case (a) of Table A14–2 gives the following formula for the amount of deflection of a cantilever for a given applied force:

[latex]y=P L^{3} / 3 E I[/latex]
Solve for the force as a function of deflection:
[latex]P=3 E I y / L^{3}[/latex]

Appendix 3 gives the modulus of elasticity for steel to be [latex]E[/latex] = 207 GPa. The moment of inertia, [latex]I[/latex], for the spring cross section is found from

[latex]I=b t^{3} / 12=(6.00 \mathrm{~mm})(0.80 \mathrm{~mm})^{3} / 12=0.256 \mathrm{~mm}^{4}[/latex]
Then the force on the spring when the deflection [latex]y[/latex] is [latex]3.0 \mathrm{~mm}[/latex] is
[latex]P=\frac{3\left(207 imes 10^{9} \mathrm{~N} / \mathrm{m}^{2} ight)\left(0.256 \mathrm{~mm}^{4} ight)(3.0 \mathrm{~mm})}{(65 \mathrm{~mm})^{3}} \frac{\left(1.0 \mathrm{~m}^{2} ight)}{\left(10^{6} \mathrm{~mm}^{2} ight)}=1.74 \mathrm{~N}[/latex]
The bending moment at the support is
[latex]M=P \cdot L=(1.74 \mathrm{~N})(65 \mathrm{~mm})=113 \mathrm{~N} \cdot \mathrm{mm}[/latex]
The bending stress at point [latex]A[/latex] caused by this moment is
[latex]\sigma=\frac{M c}{I}=\frac{(113 \mathrm{~N} \cdot \mathrm{mm})(0.40 \mathrm{~mm})}{0.256 \mathrm{~mm}^{4}}=176 \mathrm{~N} / \mathrm{mm}^{2}=176 \mathrm{MPa}[/latex]

This is the lowest stress that the spring sees in service, and therefore [latex]\sigma_{\min }=176 \mathrm{MPa}[/latex].
Because the force on the spring is proportional to the deflection, the force exerted when the deflection is [latex]8.0 \mathrm{~mm}[/latex] is
[latex]P=(1.74 \mathrm{~N})(8.0 \mathrm{~mm}) /(3.0 \mathrm{~mm})=4.63 \mathrm{~N}[/latex]
The bending moment is
[latex]M=P \cdot L=(4.63 \mathrm{~N})(65 \mathrm{~mm})=301 \mathrm{~N} \cdot \mathrm{mm}[/latex]
The bending stress at point [latex]A[/latex] is
[latex]\sigma=\frac{M c}{I}=\frac{(301 \mathrm{~N} \cdot \mathrm{mm})(0.40 \mathrm{~mm})}{0.256 \mathrm{~mm}^{4}}=470 \mathrm{~N} / \mathrm{mm}^{2}=470 \mathrm{MPa}[/latex]
This is the maximum stress that the spring sees, and therefore [latex]\sigma_{\max }=470 \mathrm{MPa}[/latex].
Now the mean stress can be computed:
[latex]\sigma_{m}=\left(\sigma_{\max }+\sigma_{\min } ight) / 2=(470+176) / 2=323 \mathrm{MPa}[/latex]
Finally, the alternating stress is
[latex]\sigma_{a}=\left(\sigma_{\max }-\sigma_{\min } ight) / 2=(470-176) / 2=147 \mathrm{MPa}[/latex]
The stress ratio is found using Equation (5-3):
Stress ratio [latex]R=\frac{ ext { minimum stress }}{ ext { maximum stress }}=\frac{\sigma_{\min }}{\sigma_{\max }}=\frac{176 \mathrm{MPa}}{470 \mathrm{MPa}}=0.37[/latex]

Comments: The sketch of stress versus time shown in Figure 5–4(a) illustrates the form of the fluctuating stress on the spring. In Section 5–8, you will see how to design parts subjected to this kind of stress.

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

TABLE A14–2 Beam-Deflection Formulas for Cantilevers
At [latex]B[/latex] at end: [latex]y_{B}=y_{\max }=\frac{-P L^{3}}{3 E I}[/latex] Between [latex]A[/latex] and [latex]B[/latex]: [latex]y=\frac{-P x^{2}}{6 E I}(3 L-x)[/latex]
At [latex]B[/latex] at load: [latex]y_{B}=\frac{-P a^{3}}{3 E I}[/latex] At [latex]C[/latex] at end: [latex]y_{C}=y_{\max }=\frac{-P a^{2}}{6 E I}(3 L-a)[/latex] Between [latex]A[/latex] and [latex]B[/latex] : [latex]y=\frac{-P x^{2}}{6 E I}(3 a-x)[/latex] Between [latex]B[/latex] and [latex]C[/latex] : [latex]y=\frac{-P a^{2}}{6 E I}(3 x-a)[/latex]
[latex]W= ext { total load }=w L[/latex] At [latex]B[/latex] at end: [latex]y_{B}=y_{\max }=\frac{-W L^{3}}{8 E I}[/latex] Between [latex]A[/latex] and [latex]B[/latex] : [latex]y=\frac{-W x^{2}}{24 E I L}\left[2 L^{2}+(2 L-x)^{2} ight][/latex]
[latex]M_{B}=[/latex] concentrated moment at end At [latex]B[/latex] at end: [latex]y_{B}=y_{\max }=\frac{-M_{B} L^{2}}{2 E I}[/latex] Between [latex]A[/latex] and [latex]B[/latex] : [latex]y=\frac{-M_{B} x^{2}}{2 E I}[/latex]

Question 5.1: For the flat steel spring shown in Figure 5–6, compute the m...

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