Page 27 - Marks Calculation for Machine Design

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January 4, 2005
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Brown.cls
Brown˙C01
FUNDAMENTAL LOADINGS
Note that Eq. (1.7) is valid only in the region up to the proportional limit as it derives
from Eq. (1.3) (Hooke’s law), where the axial stress (σ) is substituted from Eq. (1.1) and 9
the axial strain (ε) is substituted from Eq. (1.2), then rearranged to give the elongation (δ)
given in Eq. (1.7). This algebraic process is shown in Eq. (1.8).
P δ PL
σ = Eε → = E → δ = (1.8)
A L AE
As stated earlier, if the forces acting on the bar were in opposite direction, then the bar
would be loaded in compression, producing a compressive normal stress and a shortening
of the bar.
U.S. Customary SI/Metric
Example 5. Calculate the change in length of Example 5. Calculate the change in length of
a circular steel rod of radius (r) and length (L) a circular steel rod of radius (r) and length (L)
loaded axially in tension by forces (P), where loaded axially in tension by forces (P), where
P = 15 kip = 15,000 lb F = 67.5 kN = 67,500 N
r = 1.5 in r = 4cm = 0.04 m
L = 6ft L = 2m
6
2
2
9
E = 30 × 10 lb/in (steel) E = 207 × 10 N/m (steel)
solution solution
Step 1. Calculate the cross-sectional area (A) Step 1. Calculate the cross-sectional area (A)
of the rod. of the rod.
2
2
2
2
A = πr = π(1.5in) = 7in 2 A = πr = π(0.04 m) = 0.005 m 2
Step 2. Substitute the force (P), the length Step 2. Substitute the force (P), the length
(L), the area (A), and the modulus of elasticity (L), the area (A), and the modulus of elasticity
(E) in Eq. (1.7) to give the elongation (δ) as (E) into Eq. (1.7) to give the elongation (δ) as
PL (15,000 lb)(6ft) PL (67,500 N)(2m)
δ = = δ = =
2
2
9
2
2
6
AE (7in )(30 × 10 lb/in ) AE (0.005 m )(207 × 10 N/m )
90,000 lb · ft 135,000 N · m
= 6 =
210 × 10 lb 1.035 × 10 N
9
= 4.3 × 10 −4 ft × 12 in/ft = 1.3 × 10 −4 m × 1,000 mm/m
= 0.005 in = 0.13 mm
Example 6. Calculate the compressive axial Example 6. Calculate the compressive axial
forces (P) required to shorten an aluminum forces (P) required to shorten an aluminum
square bar with sides (a) and length (L) by an square bar with sides (a) and length (L) by an
amount (δ), where amount (δ), where
δ = 0.03 in = 0.0025 ft δ = 0.7 mm = 0.0007 m
a = 3in a = 8cm = 0.08 m
L = 3ft L = 1m
9
2
2
6
E = 11 × 10 lb/in (aluminum) E = 71 ×10 N/m (aluminum)
solution solution
Step 1. Calculate the cross-sectional area (A) Step 1. Calculate the cross-sectional area (A)
of the bar. of the rod.
2
2
2
2
A = a = (3in) = 9in 2 A = a = (0.08 m) = 0.0064 m 2