Residential Construction Academy: House Wiring (MindTap Course List)
4th Edition
ISBN: 9781285852225
Author: Gregory W Fletcher
Publisher: Cengage Learning
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Question
Chapter 7, Problem 21RQ
To determine
The definition of intersystem bonding termination.
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A double-angle shape, 2L7 × 4 × 5/8, is used as a tension member. The two angles are connected to a gusset plate with
7/8-inch-diameter bolts through the 7-inch legs, as shown in the figure below. A572 Grade 50 steel is used: Fy
Fu = 65 ksi. Suppose that t = 5/8 in.
= 50 ksi,
For L7 x 4 x 5/8: Ag = 6.5 in.², x = 0.958 in.
21/2"
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11/2"
2L7 x 4 x t
a. Compute the design strength.
(Express your answer to three significant figures.)
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b. Compute the allowable strength.
(Express your answer to three significant figures.)
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Sit
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I really need help on b
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Example 6
For the exterior 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 at exterior support)
D= 6.5 kN/m²
L= 5.0 kN/m²
تفكر
وکھل
flat slap
ما
لا يوجد bon حامل .
3000
1000
5000
160
+ 2000+
+2000+
5000
2608
300
2000
Drop Panal
السعف
Chapter 7 Solutions
Residential Construction Academy: House Wiring (MindTap Course List)
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- Example 4 For the transverse interior frame (Frame C) of the flat plate floor with edge beams shown in Figure, by using the Direct Design Method, find: 1) Longitudinal distribution of total static moment at factored loads. 2) Lateral distribution of moment at interior panel (column and middle strip moments atnegative and positive moments). 3) Lateral distribution of moment at exterior panel (column and middle strip moments atnegative and positive moments). Plat 5000-5000 5000 -Frame C لا بوجود deen 0009 0009 Slab thickness = 180 mm, d = 150 mm q₁ = 16.0 kN/m² All edge beams = 250x 500 mm All columns = 500x 500 mm 6000arrow_forwards الله + 600 2 Example 5 For the exterior longitudinal frame (Frame B) of the flat plate 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) Slab thickness = 175 mm, d=140 mm qu=14.0 kN/m² All columns = 600x 400 mm 916 *5000*5000*5000* B Sinter line 16400- 6400 -6400-arrow_forwardExample 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 mmarrow_forward
- 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 marrow_forwardExample 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- 300-1000arrow_forwardDetermine 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_forward
- Consider 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_forward00 N 50° W NAZ 310 Length & plane Survey! E Х A (9 13arrow_forward
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