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

M (14,400 lb ⋅ ft)(12 in./ft)
∴ S ≥ =
σ
1,800 psi
allow
= 96.0 in.
Section Modulus for a Rectangular Section
For a solid rectangular section with width b and height h, the following formula can be
derived for the section modulus:
Iz
bh /12 bh
S = = =
c h /2 6
3
3 2
The aspect ratio specified for the beam in this problem is h/b = 2; therefore, h = 2b.
Substituting this requirement into the section modulus formula gives
2 2
bh b(2 b)
4
S = = = b =
6 6 6
2
b
3
3 3
The minimum required beam width can now be determined:
2
b
3
≥ 96.0 in. ∴b
≥ 5.24 in.
Ans.
3 3
ExAmpLE 8.6
The beam shown will be constructed from a standard
steel W shape with an allowable bending stress of 30 ksi.
(a) Develop a list of acceptable shapes that could be
used for this beam. Include the most economical
W8, W10, W12, W14, W16, and W18 shapes on the
list of possibilities.
(b) Select the most economical W shape for the beam.
55 kip·ft
A
30 kips
2.5 kips/ft
B
1 kip/ft
5 ft 7 ft
4 ft
C
15 kips
D
Plan the Solution
By the graphical method presented in Section 7.3, the shear-force and bendingmoment
diagrams for the beam and loading will be constructed at the outset. With the use
of the maximum internal bending moment and the specified allowable bending stress, the
required section modulus can be determined from the flexure formula [Equation (8.10)].
Acceptable standard steel W shapes will be selected from Appendix B, and the lightest of
those shapes will be chosen as the most economical shape for this application.
SolutioN
Support Reactions
An FBD of the beam is shown. From this FBD, the equilibrium equations can be written as
follows:
Σ Fy = Ay + Cy
− 30 kips −15 kips − 30 kips − 4kips = 0
Σ M = (30 kips)(7 ft) + (30 kips)(6 ft) − (4 kips)(2 ft) − (15kips)(4 ft) + 55kip⋅ft − A (12ft) = 0
C
y
267