Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

The total elongation of the bar is the sum of the elongations of the two segments:
u = δ + δ = 0.0250 in. + 0.0458 in. = 0.0708 in. ↓ Ans.
A 1 2
Note: If the weight of the bar had not been neglected, the internal force F in both uniformwidth
segment (1) and tapered segment (2) would not have been constant and Equation (5.5)
would be required for both segments. To include the weight of the bar in the analysis, a function
expressing the change in internal force as a function of the vertical position y should be derived
for each segment. The internal force F at any position y is the sum of a constant force equal to
P and a varying force equal to the self-weight of the axial member below position y. The force
due to self-weight will be a function that expresses the volume of the bar below any position
y, multiplied by the specific weight of the material that the bar is made of. Since the internal
force F varies with y, it must be included inside the integral in Equation (5.5).
mecmovies
ExERcISES
m5.1 Use the axial deformation equation for three introductory
problems.
m5.2 Apply the axial deformation concept to compound axial
members.
FIGURE m5.2
FIGURE m5.1
pRoBLEmS
p5.1 A steel [E = 200 GPa] rod with a circular cross section is
15 m long. Determine the minimum diameter required if the rod
must transmit a tensile force of 300 kN without exceeding an allowable
stress of 250 MPa or stretching more than 10 mm.
P
b
b
3
P
p5.2 A rectangular bar of length L has a slot in the central half of its
length, as shown in Figure P5.2. The bar has width b, thickness t, and
elastic modulus E. The slot has width b/3. If L = 400 mm, b = 45 mm,
L
4
FIGURE p5.2
L
2
L
4
92