A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2is subjected to an external tensile
load. If the stresses sustained by the fiber and matrix phases are 156 MPa and 2.75 MPa, respectively, the force sustained by the fiber phase is 74,000 Nand the total longitudinal strain is1.25x 10-3, determine (a)the force sustained by the matrix phase, (b)the modulus of elasticity of the composite material in the longitudinal direction, and (c)the moduli of elasticity for the fiber and matrix phases.
1 answer:
Answer:
(a) The force sustained by the matrix phase is 1802.35 N
(b) The modulus of elasticity of the composite material in the longitudinal direction Ed is 53.7 GPa
(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively
Explanation:
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Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*/2*c^4
∅=2TL/G**c^4
c=∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*/2*c^4)]*c
г =[2T/(J**c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm
Answer:
(a) The Final Temperature is 315.25 K.
(b) The amount of mass that has entered 0.5742 Kg.
(c) The work done is 56.52 kJ.
(d) The entrophy generation is 0.0398 kJ/kgK.
Explanation:
Explanation is in the following attachments.
