EBK ENGINEERING FUNDAMENTALS: AN INTROD
5th Edition
ISBN: 8220100543401
Author: MOAVENI
Publisher: CENGAGE L
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Chapter 6.2, Problem 3BYG
To determine
Find the name at least three prefixes and symbols of the decimal multiples of SI units.
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An L6 × 4 × 5/8 tension member of A36 steel is connected to a gusset plate with 1-inch-diameter bolts, as shown in the
figure below. It is subjected to the following service loads: dead load = 40 kips, live load = 100 kips, and wind load
= 45 kips. Use the equation for U: U = 1 −
For A36 steel: Fy = 36 ksi, F = 58 ksi.
x
l
For L6 × 4×5/8: Ag = 5.86 in.², x = 1.03 in.
21/4"
L6 × 4 × 5/8
11/2"
21/2"
11/2"
a. Determine whether this member is adequate using LRFD.
-Select-
What is the design strength for LFRD?
(Express your answer to three significant figures.)
Φι Ρη -
=
kips
Which AISC load combination controls?
-Select-
What is the controlling AISC load combination?
(Express your answer to three significant figures.)
Pu
=
kips
b. Determine whether this member is adequate using ASD.
-Select-
What is the allowable strength for ASD?
(Express your answer to three significant figures.)
Pn
Sit
kips
Which ASD load combination controls?
-Select-
What is the controlling ASD load combination?…
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"
оо
11/2"
2L7 x 4 x t
a. Compute the design strength.
(Express your answer to three significant figures.)
ФЕРП
=
kips
b. Compute the allowable strength.
(Express your answer to three significant figures.)
'n
Sit
=
kips
I really need help on b
Chapter 6 Solutions
EBK ENGINEERING FUNDAMENTALS: AN INTROD
Ch. 6.1 - Prob. 1BYGCh. 6.1 - Prob. 2BYGCh. 6.1 - Prob. 3BYGCh. 6.1 - Prob. 4BYGCh. 6.1 - Prob. BYGVCh. 6.2 - Prob. 1BYGCh. 6.2 - Prob. 2BYGCh. 6.2 - Prob. 3BYGCh. 6.2 - Prob. 4BYGCh. 6.2 - Prob. 5BYG
Ch. 6.2 - Prob. 6BYGCh. 6.2 - Prob. BYGVCh. 6.3 - Prob. 1BYGCh. 6.3 - Prob. 2BYGCh. 6.3 - Prob. 3BYGCh. 6.3 - Prob. 4BYGCh. 6.3 - Prob. BYGVCh. 6.5 - Prob. 1BYGCh. 6.5 - Prob. 2BYGCh. 6.5 - Prob. 3BYGCh. 6.5 - Prob. 4BYGCh. 6.5 - Prob. BYGVCh. 6.6 - Prob. 1BYGCh. 6.6 - Prob. 2BYGCh. 6.6 - Prob. 3BYGCh. 6.6 - Prob. BYGVCh. 6 - Prob. 1PCh. 6 - Prob. 2PCh. 6 - Prob. 3PCh. 6 - Prob. 4PCh. 6 - Prob. 5PCh. 6 - Prob. 6PCh. 6 - Prob. 7PCh. 6 - Prob. 8PCh. 6 - Prob. 9PCh. 6 - Prob. 10PCh. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - Prob. 17PCh. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - Prob. 22PCh. 6 - Prob. 23PCh. 6 - The air resistance to the motion of a vehicle is...Ch. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Prob. 27PCh. 6 - Prob. 28PCh. 6 - Prob. 29PCh. 6 - Prob. 30PCh. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - The calorie is defined as the amount of heat...Ch. 6 - Prob. 35PCh. 6 - Prob. 36PCh. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - For the fin equation described in Problem 6.25, if...Ch. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Prob. 44PCh. 6 - Prob. 45PCh. 6 - Prob. 46PCh. 6 - Prob. 47PCh. 6 - Prob. 48PCh. 6 - Prob. 49PCh. 6 - Prob. 50PCh. 6 - Prob. 51P
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- WU 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 السعفarrow_forwardExample 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_forward
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