Business Driven Information Systems
6th Edition
ISBN: 9781260004717
Author: Paige Baltzan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 2, Problem 5OCQ
To determine
The different metrics that a personal trainer robot for a fitness club can provide to a customer.
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6. A simply supported beam is subjected to uniformly distributed loads: a service dead load
WD =
5 kips/ft and a service live load w₁ =7 kips/ft. Use the load combination w₁ = 1.2WD +
1.6WL. Note that the shear force is expressed as V₂ = w₁ (-/-- x) where x is the distance from a
support. (65pts total)
The beam also has the following properties.
f = 4ksi,
fy = fyt = 60 ksi,
b = 18 in,
h = 26 in,
d = 24.5 in,
20ft
No. 4 bars, U-stirrups used, Normalweight concrete
(1) Calculate the required spacing of stirrups. (15pts)
(2) Calculate the maximum spacing. Also, determine the location along the beam where the
maximum spacing can be applied (i.e., the distance from support). (10pts)
(3) Show the stirrup designs along with the shear force diagram. Indicate the numbers and
spacings of stirrups. A transition between the required spacing and the maximum spacing is
unnecessary. (15pts)
(4) Assume the live load decreases, and, as a result, the factored shear force is 30 kips. Discuss
whether…
Mechanical engineering, Don't use chatgpt.
Strict warning.
5. A singly reinforced beam has a width b = 16 in., a height h = 20 in., and
an effective depth d = 18.5 in. The beam is reinforced with six No. 8 bars.
The beam is subjected to a positive bending moment, causing the bottom of
the beam to experience tension. (65pts total)
The material properties are as follows:
f = 5 ksi, fy = 60 ksi
d
888
Six no. 8 bars
(1) Find the bending moment [kips-in] that can cause tension cracks. Use the following
parameters regarding the equivalent transformed section. Do not consider load and resistance
factors. (15pts)
- Neutral axis: 10.7 in. from the top face of beam
- Moment of inertia: 12,550 in² around the neutral axis
- Modulus of rupture: fƒ„ = 7.5√ƒ! [psi] (i.e., the concrete tensile stress for crack initiation)
(2) Find the service moment strength [kips-in]. The allowable stress of concrete and steel is
0.45f, and 0.5fy, respectively. Do not consider load and resistance factors. (15pts)
(3) Calculate the nominal moment strength M, and the design…
Chapter 2 Solutions
Business Driven Information Systems
Ch. 2 - Prob. 1OCQCh. 2 - Prob. 2OCQCh. 2 - Prob. 3OCQCh. 2 - Prob. 4OCQCh. 2 - Prob. 5OCQCh. 2 - Prob. 6OCQCh. 2 - Prob. 1RQCh. 2 - Prob. 2RQCh. 2 - Prob. 3RQCh. 2 - Prob. 4RQ
Ch. 2 - Prob. 5RQCh. 2 - Prob. 6RQCh. 2 - Prob. 7RQCh. 2 - Prob. 8RQCh. 2 - Prob. 9RQCh. 2 - Prob. 10RQCh. 2 - Prob. 11RQCh. 2 - Prob. 12RQCh. 2 - Prob. 13RQCh. 2 - Prob. 14RQCh. 2 - Prob. 15RQCh. 2 - Prob. 16RQCh. 2 - Prob. 17RQCh. 2 - Prob. 1CCOCh. 2 - Prob. 2CCOCh. 2 - Prob. 3CCOCh. 2 - Prob. 4CCOCh. 2 - Prob. 5CCOCh. 2 - Prob. 6CCOCh. 2 - Prob. 1CCTCh. 2 - Prob. 2CCTCh. 2 - Prob. 3CCTCh. 2 - Prob. 1CBTCh. 2 - Prob. 2CBTCh. 2 - Prob. 3CBTCh. 2 - Prob. 4CBTCh. 2 - Prob. 5CBTCh. 2 - Prob. 6CBTCh. 2 - Prob. 7CBTCh. 2 - Prob. 8CBTCh. 2 - Prob. PIAYKBPCh. 2 - Prob. PIIAYKBPCh. 2 - Prob. PIIIAYKBPCh. 2 - Prob. PIVAYKBPCh. 2 - Prob. PVAYKBPCh. 2 - Prob. PVIAYKBPCh. 2 - Prob. PVIIAYKBPCh. 2 - Prob. PVIIIAYKBPCh. 2 - Prob. PIXAYKBPCh. 2 - Prob. PXAYKBPCh. 2 - Prob. PXIAYKBP
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