Find the maximum value of S 0 if a stress σx = 2S, σy = –3S, and τxy = 4S are applied to a 60° lamina of graphite/epoxy. Use Tsai–Wu failure theory. Use the properties of a unidirectional graphite/epoxy lamina from Table 2.1.

Question

Find the maximum value of S > 0 if a stress [latex]σ_x = 2S, σ_y = –3S, and τ_{xy }= 4S[/latex] are applied to a 60° lamina of graphite/epoxy. Use Tsai–Wu failure theory. Use the properties of a unidirectional graphite/epoxy lamina from Table 2.1.

TABLE 2.1
Typical Mechanical Properties of a Unidirectional Lamina (SI System of Units)

Property	Symbol	Units	Glass/ epoxy	Boron/ epoxy	Graphite/ epoxy
Fiber volume fraction	[latex]V_f[/latex]		0.45	0.50	0.70
Longitudinal elastic modulus	[latex]E_1[/latex]	GPa	38.6	204	181
Transverse elastic modulus	[latex]E_2[/latex]	GPa	8.27	18.50	10.30
Major Poisson’s ratio	[latex]V_{12}[/latex]		0.26	0.23	0.28
Shear modulus	[latex]G_{12}[/latex]	GPa	4.14	5.59	7.17
Ultimate longitudinal tensile strength	[latex](\sigma_1^T)_{ult}[/latex]	MPa	1062	1260	1500
Ultimate longitudinal compressive strength	[latex](\sigma_1^C)_{ult}[/latex]	MPa	610	2500	1500
Ultimate transverse tensile strength	[latex](\sigma_2^T)_{ult}[/latex]	MPa	31	61	40
Ultimate transverse compressive strength	[latex](\sigma_2^C)_{ult}[/latex]	MPa	118	202	246
Ultimate in-plane shear strength	[latex]( au_{12})_{ult}[/latex]	MPa	72	67	68
Longitudinal coefficient of thermal expansion	[latex]\alpha_1[/latex]	μm/m/°C	8.6	6.1	0.02
Transverse coefficient of thermal expansion	[latex]\alpha_2[/latex]	μm/m/°C	22.1	30.3	22.5
Longitudinal coefficient of moisture expansion	[latex]\beta_1[/latex]	m/m/kg/kg	0.00	0.00	0.00
Transverse coefficient of moisture expansion	[latex]\beta_2[/latex]	m/m/kg/kg	0.60	0.60	0.60

Source: Tsai, S.W. and Hahn, H.T., Introduction to Composite Materials, CRC Press, Boca Raton, FL, Table 1.7, p. 19; Table 7.1, p. 292; Table 8.3, p. 344. Reprinted with permission.

TABLE 1.7
Chemical Composition of E-Glass and S-Glass Fibers

Material	% Weight
Material	E-Glass	S-Glass
Silicon oxide	54	64
Aluminum oxide	15	25
Calcium oxide	17	0.01
Magnesium oxide	4.5	10
Boron oxide	8	0.01
Others	1.5	0.8

Accepted Answer

From Example 2.13,

[latex]\sigma_1=1.714S, \ \space \ \sigma_2=-2.714S, \ \space \ au_{12}=-4.165S.[/latex]

From Equations (2.155), (2.156), (2.159), (2.160), (2.163), and (2.164),

[latex] H_{1}=\frac{1}{1500 imes 10^{6}}-\frac{1}{1500 imes 10^{6}}=0 Pa ^{-1}, \ \space \ H_{2}=\frac{1}{40 imes 10^{6}}-\frac{1}{246 imes 10^{6}}=2.093 imes 10^{-8} Pa ^{-1}, \ \space \ H_6=0 Pa^{-1}, \ \space \ H_{11}=\frac{1}{\left(1500 imes 10^{6} ight)\left(1500 imes 10^{6} ight)}=4.4444 imes 10^{-19} Pa ^{-2}, \ \space \H_{22}=\frac{1}{\left(40 imes 10^{6} ight)\left(246 imes 10^{6} ight)}=1.0162 imes 10^{-16} Pa ^{-2}, \ \space \ H_{66}=\frac{1}{\left(68 imes 10^{6} ight)^{2}}=2.1626 imes 10^{-16} Pa ^{-2} .[/latex]

Using the Mises–Hencky criterion for evaluation of [latex]H_{12}[/latex], (Equation 2.165c),

[latex] H_{12}=-\frac{1}{2} \sqrt{\frac{1}{\left(1500 imes 10^{6} ight)\left(1500 imes 10^{6} ight)\left(40 imes 10^{6} ight)\left(246 imes 10^{6} ight)}}=3.360 imes 10^{-18} Pa ^{-2} . [/latex]

Substituting these values in Equation (2.152), we obtain

[latex] (0)(1.714 S)+\left(2.093 imes 10^{-8} ight)(-2.714 S) \ \space \ +(0)(-4.165 S)+\left(4.444 imes 10^{-19} ight)(1.714 S)^{2} \ \space \ +\left(1.0162 imes 10^{-16} ight)(-2.714 S)^{2}+\left(2.1626 imes 10^{-16} ight)(-4.165 S)^{2} \ \space \ +2\left(-3.360 imes 10^{-18} ight)(1.714 S)(-2.714 S)<1, [/latex]

or

S < 22.39 MPa.

If one uses the other two empirical criteria for [latex]H_{12}[/latex], per Equation (2.171), this yields

[latex] S<22.49 MPa ext { for } H_{12}=-\frac{1}{2\left(\sigma_{1}^{T} ight)_{u l t}^{2}}, \ \space \ S<22.49 MPa ext { for } H_{12}=-\frac{1}{2} \frac{1}{\left(\sigma_{1}^{T} ight)_{u l t}\left(\sigma_{1}^{C} ight)_{u l t}}. [/latex]

Summarizing the four failure theories for the same stress state, the value of S obtained is

S = 16.33 (maximum stress failure theory)
S = 16.33 (maximum strain failure theory)
S = 10.94 (Tsai–Hill failure theory)
S = 16.06 (modified Tsai–Hill failure theory)
S = 22.39 (Tsai–Wu failure theory)

_______________________________________

(2.152): [latex] H_{1} \sigma_{1}+H_{2} \sigma_{2}+H_{6} au_{12}+H_{11} \sigma_{1}^{2}+H_{22} \sigma_{2}^{2}+H_{66} au_{12}^{2}+2 H_{12} \sigma_{1} \sigma_{2}<1 [/latex]

(2.155): [latex] H_{1}=\frac{1}{\left(\sigma_{1}^{T} ight)_{u l t}}-\frac{1}{\left(\sigma_{1}^{C} ight)_{u l t}} [/latex]

(2.156): [latex] H_{11}=\frac{1}{\left(\sigma_{1}^{T} ight)_{u l t}\left(\sigma_{1}^{C} ight)_{u l t}} [/latex]

(2.159): [latex] H_{2}=\frac{1}{\left(\sigma_{2}^{T} ight)_{u l t}}-\frac{1}{\left(\sigma_{2}^{C} ight)_{u l t}} [/latex]

(2.160): [latex] H_{22}=\frac{1}{\left(\sigma_{2}^{T} ight)_{u l t}\left(\sigma_{2}^{C} ight)_{u l t}} [/latex]

(2.163): [latex]H_6=0[/latex]

(2.164): [latex] H_{66}=\frac{1}{\left( au_{12} ight)_{u l t}^{2}} [/latex]

(2.165): [latex] \left(H_{1}+H_{2} ight) \sigma+\left(H_{11}+H_{22}+2 H_{12} ight) \sigma^{2}=1 [/latex]

(2.171):
[latex] H_{12}=-\frac{1}{2\left(\sigma_{1}^{T} ight)_{u l t}^{2}}, ext { per Tsai-Hill failure theory }{ }^{8} \ \space \ H_{12}=-\frac{1}{2\left(\sigma_{1}^{T} ight)_{u l t}\left(\sigma_{1}^{C} ight)_{u l t}}, ext { per Hoffman criterion }{ }^{10} \ \space \ H_{12}=-\frac{1}{2} \sqrt{\frac{1}{\left(\sigma_{1}^{T} ight)_{u l t}\left(\sigma_{1}^{C} ight)_{u l t}\left(\sigma_{2}^{T} ight)_{u l t}\left(\sigma_{2}^{C} ight)_{u l t}}} ext {, per Mises-Hencky criterion. }{ }^{11}[/latex]

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