(a) When a constant load is applied to a bar with a constant cross section, as shown in Figure 6.5, the amount by which the end of the bar will deflect can be determined from the relationship PL d = PL d = AE (N)(m) (m²)E 6.7 AE Solving for the units of E leads to N/m² (called newton per squared meter or force per unit area). (b) The heat transfer rate through a solid material is governed by Fourier's law: - T2 9 = kA L 6.8 where q = heat transfer rate k = thermal conductivity of the solid material in watts per meter degree Celsius, W/m · °C A = area in m² (a) (b) FIGURE 6.5 (a) The bar in Example 6.6 and (b) heat transfer through a solid material. T, - T, = temperature difference, °C L = thickness of the material, m What is the appropriate unit for the heat transfer rate q? Substituting for the units of k, A, T, , T, , and L in Equation 6.8, we have where d = end deflection of the bar in meter (m) P = applied load in newton (N) L = length of the bar in meter (m) A = cross-sectional area of the bar (m²) E = modulus of elasticity of the material What are the units for modulus of elasticity? W q= kA- L = W m Im.c(m? From this, you can see that the appropriate SI unit for the heat transfer rate is the watt. For Equation (6.7) to be dimensionally homogeneous, the units on the left- hand side of the equation must equal the units on the right-hand side. This equality requires the modulus of elasticity to have the units of N/m?, as
(a) When a constant load is applied to a bar with a constant cross section, as shown in Figure 6.5, the amount by which the end of the bar will deflect can be determined from the relationship PL d = PL d = AE (N)(m) (m²)E 6.7 AE Solving for the units of E leads to N/m² (called newton per squared meter or force per unit area). (b) The heat transfer rate through a solid material is governed by Fourier's law: - T2 9 = kA L 6.8 where q = heat transfer rate k = thermal conductivity of the solid material in watts per meter degree Celsius, W/m · °C A = area in m² (a) (b) FIGURE 6.5 (a) The bar in Example 6.6 and (b) heat transfer through a solid material. T, - T, = temperature difference, °C L = thickness of the material, m What is the appropriate unit for the heat transfer rate q? Substituting for the units of k, A, T, , T, , and L in Equation 6.8, we have where d = end deflection of the bar in meter (m) P = applied load in newton (N) L = length of the bar in meter (m) A = cross-sectional area of the bar (m²) E = modulus of elasticity of the material What are the units for modulus of elasticity? W q= kA- L = W m Im.c(m? From this, you can see that the appropriate SI unit for the heat transfer rate is the watt. For Equation (6.7) to be dimensionally homogeneous, the units on the left- hand side of the equation must equal the units on the right-hand side. This equality requires the modulus of elasticity to have the units of N/m?, as
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Saeed Moaveni
Chapter7: Length And Length-related Variables In Engineering
Section: Chapter Questions
Problem 43P: A 10 cm long rectangular bar (when subjected to a tensile load) deforms by 0.1 mm. Calculate the...
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For Example 6.6 part (a), if the units of d, P, L , and A are given in in., lbf , inches, and in2, respectively, what are the units for modulus of elasticity E ?
Transcribed Image Text:(a) When a constant load is applied to a bar with a constant cross section,
as shown in Figure 6.5, the amount by which the end of the bar will
deflect can be determined from the relationship
PL
d =
PL
d =
AE
(N)(m)
(m²)E
6.7
AE
Solving for the units of E leads to N/m² (called newton per squared meter
or force per unit area).
(b) The heat transfer rate through a solid material is governed by Fourier's
law:
- T2
9 = kA
L
6.8
where
q = heat transfer rate
k = thermal conductivity of the solid material in watts per meter
degree Celsius, W/m · °C
A = area in m²
(a)
(b)
FIGURE 6.5
(a) The bar in Example 6.6 and (b) heat transfer
through a solid material.
T, - T, = temperature difference, °C
L = thickness of the material, m
What is the appropriate unit for the heat transfer rate q?
Substituting for the units of k, A, T, , T, , and L in Equation 6.8, we have
where
d = end deflection of the bar in meter (m)
P = applied load in newton (N)
L = length of the bar in meter (m)
A = cross-sectional area of the bar (m²)
E = modulus of elasticity of the material
What are the units for modulus of elasticity?
W
q= kA-
L
= W
m
Im.c(m?
From this, you can see that the appropriate SI unit for the heat transfer rate
is the watt.
For Equation (6.7) to be dimensionally homogeneous, the units on the left-
hand side of the equation must equal the units on the right-hand side. This
equality requires the modulus of elasticity to have the units of N/m?, as
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