ENGINEERING FUNDAMENTALS
6th Edition
ISBN: 9781337705011
Author: MOAVENI
Publisher: CENGAGE L
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Chapter 2.5, Problem 1BYG
To determine
Explain the importance to get involve with an engineering organization.
Expert Solution & Answer
Explanation of Solution
An Engineering organization includes various good reasons to involve in it. Few reasons are listed below:
- Opportunity to learn through short courses, conferences, and seminars,
- Plant tour participation,
- Design Development,
- Applicable to practical life.
The primary purpose of an engineering organization includes the following benefits,
- 1. To share the new creative concepts and the Research and development findings, international level conferences can be held.
- 2. Technical journal, reports, magazines and book are developed and published in order to keep an engineer as up-to-date in particular fields.
- 3. The recent technology development has been offered as a short course to develop the technical knowledge in their respective fields.
- 4. An engineer can make sure of the public safety by providing the standard codes using the engineering designs.
- 5. Engineering organization can provide the best networking
mechanism through which the strong relationship is made along with various institutions and organizations.
Conclusion:
Thus, the importance to get involve with an engineering organization has been explained.
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H.W.5: A beam section is limited to a width of 300 mm, and total depth h=500 mm, and has to resist
a factored moment 405 kN.m.Calculate the required reinforcement, given f'c=28 MPa,
fy=420 MPa.
(As=6425mm, As'=2425 mm Ans)
H.W.1: A rectangular beam has a width of 300 mm, and effective depth d=570 mm to centroid of
tension steel bars. Tension steel reinforcement consist of 6628mm in two rows,
compression reinforcement of 2022mm. Calculate the design moment strength of the
beam, where f'c=28 MPa, fy=420 MPa.
(Mu=822 kN.m Ans)
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- H.W.4: A beam section is limited to a width of 250 mm, and total depth h=550 mm, and has to resist a factored moment 307 kN.m.Calculate the required reinforcement, given f'c=21 MPa, fy=350 MPa, d'=65mm. (As=5425mm, As'=2022 mm Ans) HW5 A beam section is limited to a width of 300 mm and total denth h-500 mm and has to resistarrow_forwardH.W.3: A rectangular beam has a width of 400 mm, and effective depth d=700 mm to centroid of tension steel bars. Tension steel reinforcement consist of 4036mm in two rows, compression reinforcement of 2622mm. Calculate the design moment strength of the beam, where f'c=21 MPa, fy=420 MPa, d'=65mm. (Mu=927 kN.m Ans)arrow_forwardH.W.4: A beam section is limited to a width of 250 mm, and total depth h=550 mm, and has to resist a factored moment 307 kN.m.Calculate the required reinforcement, given f'c=21 MPa, fy=350 MPa, d'=65mm. (As=5425mm, As'=2422 mm Ans)arrow_forward
- H.W.6: Design the steel reinforcement for flexural for the beam shown in the fig. below. Given f'c=28 MPa, fy=420 MPa. D.L 100 kN/m L.L=200 kN 3 m 3 m 600 mm- A =? 300 mm.arrow_forwardThe tension member shown in the figure below must resist a service dead load of 60 kips and a service live load of 45 kips. Does the member have enough strength? The steel is A588: Fy = 50 ksi, F₁ = 70 ksi; and the bolts are 11/8 inches in diameter. Assume that A = An PL 38 x 72 оо a. Use LRFD. Determine the design strength and the factored load. Make a conclusion about the member. (Express your answers to three significant figures.) +Pn Pu = = kips kips -Select- b. Use ASD. Determine the allowable strength and required strength. Make a conclusion about the member. (Express your answers to three significant figures.) Ft Ae = P₁ = -Select- kips kipsarrow_forwardA single-angle tension member of A36 steel must resist a dead load of 49 kips and a live load of 84 kips. The length of the member is 18 feet, and it will be connected with a single line of 1-inch-diameter bolts, as shown in the figure below. There will be four or more bolts in this line. For the steel Fy = 36 ksi and F₁ = 58 ksi. Try the tension members given in the table below. Tension member 4, (in.) rz (in.) 7 L6 × 6 × 9.75 1.17 8 L 5 × 3 × 4.93 0.746 L 5 × 3 × 5 2.56 0.758 16 7 L5 × 3 × 3.31 0.644 16 Bolt line a. Select a single-angle tension member to resist the loads. Use LRFD. A) L 6 × 6 × B) L 5 × 3 × CL5×3× D) L 5 × 3 × -Select- 5 16 7 16 What is the required gross area? (Express your answer to three significant figures.) A₁ = in.² What is the required effective area? (Express your answer to three significant figures.) A = in.2 What is the minimum radius of gyration? (Express your answer to three significant figures.) "min = in. b. Select a single-angle tension member to…arrow_forward
- 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?…arrow_forwardA 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 = kipsarrow_forwardI really need help on barrow_forward
- 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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