Engineering Fundamentals: An Introduction to Engineering
6th Edition
ISBN: 9780357112311
Author: Saeed Moaveni
Publisher: Cengage Learning US
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Chapter 2.3, Problem 1BYG
To determine
Explain the main difference between learning in high school and college.
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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 2 Solutions
Engineering Fundamentals: An Introduction to Engineering
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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
- 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 mmarrow_forwardExample 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_forward
- 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
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