LMS Integrated for MindTap Engineering, 2 terms (12 months) Printed Access Card for Moavni's Engineering Fundamentals: An Introduction to Engineering, 5th
5th Edition
ISBN: 9781305110366
Author: Saeed Moaveni
Publisher: Cengage Learning
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Chapter 12.4, Problem BYGV
To determine
Explain the meaning of the terms duty cycle, foot candle, efficacy, CRI, CFL and LED.
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Example 8
For the longitudinal frame of the flat slab floor shown in figure, and by using
the Direct Design Method, find:
a. Longitudinal distribution of the total static moment at factored loads.
b. Lateral distribution of moment at exterior panel (column and middle
strip moments at exterior support)
qu 18.0 kN/m²
edge beams: 300×600 mm
5000 mm
CL Panel
6000
واجب
750
750-
400
099-
5000 mm
+2000+
CL Panel
1120
Drop Panal
Cobum Cop
250
احول دائري الى توسيع
احلة
$400mm
face to face
6000 mm
Example 9
For the the transverse exterior frame (Frame D) of the flat plate floor, without edge beams, shown in
Figure, and by using the Direct Design Method, find:
a. Longitudinal distribution of the total static moment at factored loads.
b. Lateral distribution of moment at interior panel (column and middle stripmoments at
negative and positive moments).
Slab thickness = 180 mm, d = 150 mm
qu= 15.0 kN/m², All columns = 400×400 mm
5.0 m-
5.0-
5.0-
نصف عرف العمود
6.0 m
Example 7
For the transverse frame of the flat slab floor shown in figure, and by
using the Direct Design Method, find:
a. Longitudinal distribution of the total static moment at factored loads.
b. Lateral distribution of moment at exterior panel (column and middle
strip moments atexterior support)
Flit
D = 7.0 kN/m²
L = 4.0 kN/m²
3000-
5000
-160
+1000+
5000
009-
300-1000
Chapter 12 Solutions
LMS Integrated for MindTap Engineering, 2 terms (12 months) Printed Access Card for Moavni's Engineering Fundamentals: An Introduction to Engineering, 5th
Ch. 12.1 - In what unit is electric current measured?Ch. 12.1 - Prob. 2BYGCh. 12.1 - Prob. 3BYGCh. 12.1 - Prob. 4BYGCh. 12.1 - Prob. 5BYGCh. 12.1 - Prob. 6BYGCh. 12.1 - Prob. BYGVCh. 12.2 - Prob. 1BYGCh. 12.2 - Prob. 2BYGCh. 12.2 - Prob. 3BYG
Ch. 12.2 - Prob. 4BYGCh. 12.2 - Prob. 5BYGCh. 12.2 - Prob. 6BYGCh. 12.2 - VocabularyState the meaning of the following...Ch. 12.4 - Prob. 1BYGCh. 12.4 - Prob. 2BYGCh. 12.4 - Prob. 3BYGCh. 12.4 - Prob. 4BYGCh. 12.4 - Prob. 5BYGCh. 12.4 - Prob. 6BYGCh. 12.4 - Prob. BYGVCh. 12 - Prob. 1PCh. 12 - Prob. 2PCh. 12 - Prob. 3PCh. 12 - Prob. 6PCh. 12 - Prob. 8PCh. 12 - Prob. 9PCh. 12 - Prob. 10PCh. 12 - Prob. 16PCh. 12 - Prob. 18PCh. 12 - Prob. 19PCh. 12 - Prob. 20PCh. 12 - Prob. 21PCh. 12 - Prob. 23PCh. 12 - Prob. 24PCh. 12 - As you know, a fuse is a safety device that is...Ch. 12 - Prob. 27PCh. 12 - Prob. 28PCh. 12 - Prob. 29PCh. 12 - Prob. 30PCh. 12 - Prob. 31P
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- Determine the amount of rebar needed for the spread footing where the dowels extend 24 inches into the column allow for 3 inches of concrete coverarrow_forwardConsider the forces acting on the handle of the wrench in Figure 1arrow_forwardThe following table gives the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) N60 1.5 6 3.0 14 4.5 14 6.0 19 17 7.5 9.0 23 The groundwater table is located at a depth of 12 m. The dry unit weight of sand from 0 to a depth of 12 m is 17.6 kN/m³. Assume the mean grain size (D 50) of the sand deposit to be about 0.8 mm. Estimate the variation of the relative density with depth for sand. Use the equation N60 (0.23 +0.06/D50) 1.7 1 Dr (%) = 9 σ'o/Pa (Enter your answers to three significant figures.) Depth (m) N60 Dr (%) 1.5 6 3.0 14 4.5 14 6.0 19 7.5 17 9.0 23 0.5 (100)arrow_forward
- 00 N 50° W NAZ 310 Length & plane Survey! E Х A (9 13arrow_forwardA cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Cone resistance, qc (MN/m²) 2.0 3.5 5.0 6.5 8.0 4.02 5.15 6.04 10.13 13.10 The average unit weight of the sand is 16.5 kN/m³. Assume moderately compressible sand and hence Q Determine the relative density at each depth using the equation below. Dr = 1 305QOCR1.8 Яс Pa 0.5 0 Pa (Enter your answers to three significant figures.) Depth (m) Dr (%) 2.0 3.5 5.0 6.5 8.0 = 1.arrow_forwardA vane shear test was conducted in a saturated soft clay, using a 100 mm x 260 mm vane. When the vane was rotated at the standard rate of 0.1°/s, the torque measured in the torque meter increased to 60 N. m, and with further rotation reduced to 35 N. m. Determine the peak and residual undrained shear strengths of the clay. (Enter your answers to three significant figures.) Peak undrained shear strength Residual undrained shear strength = kN/m² kN/m²arrow_forward
- Following is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) Neo N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 19 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 20.2 kN/m³. Estimate an average peak soil friction angle. Use the equation CN - [ 1 (o'o/Pa). 0.5 (Enter your answer to three significant figures.) $' =arrow_forwardThe beam shown in the figure below is typical for a floor system in an existing building.It needs to carry a uniform live load of 260 lb/ft and a uniform dead weight of 400 lb/ft,including its own weight. The owner wants to add a partition weighing 7 kip (live load) asshown. Assuming the added partition as live load, is the beam section adequate to safelycarry the extra live load? a. Determine the design moment capacity .b. Determine the factored applied bending moment. c. Is the beam safe and adequate for bending? Please explain your response.arrow_forwardFollowing is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 10 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 12.2 kN/m³. Use the relationship given in the equation C'N = 1 σo/Pa 0.5 to calculate the corrected penetration numbers. (Round your answers to the nearest whole number.) Depth (m) N60 (N1) 60 1.5 6 4.5 3 8 9 6 8 7.5 13 9 14arrow_forward
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