
When did an economical production method of steel become available?

The invention of economical production method of the steel.
Explanation of Solution
Steel has been the most used material in the structures for a few centuries. The production of steel was quite expensive compared to other materials. In earlier days of the nineteenth century, the American inventors started processes which can produce the steel economically. The process was named as Bessemer process. By this process, the steel production cost reduced to 82% to the previous cost. So, this is the first economical production method invention in steel according to the production cost.
Therefore, we can say that economic production of steel started in the early 19th century.
Want to see more full solutions like this?
Chapter 1 Solutions
Structural Steel Design (6th Edition)
Additional Engineering Textbook Solutions
Mechanics of Materials (10th Edition)
Degarmo's Materials And Processes In Manufacturing
Management Information Systems: Managing The Digital Firm (16th Edition)
Modern Database Management
Electric Circuits. (11th Edition)
Java: An Introduction to Problem Solving and Programming (8th Edition)
- A sample of Achilles saturated with water has a mass of 1710 g. After heating in an oven, a constant mass of 1815 g is obtained. The density of solid Achilles seeds is 2.78 g/cm3. We are asked to calculate: a) The water content and void ratio b) The porosity and specific gravity of the clayey soil c) The wet density of the clayey soil, the corresponding dry density and dry densityarrow_forward8. A prestressed concrete beam is subjected to the following stress distributions: Pi is the initial prestressing force, Pe is the effective prestressing force, M, is the bending moment due to self- weight, Ma and M, are the dead load and live load bending moment, respectively. The concrete has the following properties: fr = 6000 psi and fri = 4200 psi +250 -85 -2500 +550 Pe+ Mo+Ma+Mi P alone P₁+ Mo -2450 -3500 Stress at midspan +210 +250 P, alone Pe alone -2500 -3500 Stress at ends Using Table 22.1, evaluate whether the stresses at the center of the span and the end of the span comply with the permissible stress limits. The beam is classified as U-class. Provide justifications for each condition listed in the table. Note: Calculated stresses are to be taken from the above diagram, and permissible stresses are to be calculated using Table 22.1. Compressive stresses immediately after transfer Tensile stresses immediately after transfer Compressive stresses under sustained and total…arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. (2pts each, 10pt total) Po Pn (1) failure range H 3" 30" Ast 6 No. 10 bars = P 22" I e H 3" (4) e = e small Load path for given e Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb.arrow_forward
- No chatgpt plsarrow_forward11. The prestressed T beam shown below is pretensioned using low relaxation stress-relieved Grade 270 strands. The steel area Aps = 2.5 in². The tensile strength is fpu = 270 ksi, and the concrete compressive strength is fr = 6000 psi. (a) Calculate the nominal moment strength Mn with hr = 6 in. 22" 15" T hf (b) Since this beam is a T-beam, the nominal moment strength M₁ increases with a thicker hf. However, M, stops increasing if he reaches a value. Determine the minimum thickness hy that can achieve the maximum nominal moment strength Mr. Also, calculate the corresponding maximum nominal moment strength Mn with the computed hf.arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward
- 10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward7. Match the given strand profiles with the corresponding loading conditions for a prestressed concrete (PSC) beam. Strand profile (b) (d) (c) (a) Ꮎ Load on a beamarrow_forward4. For serviceability considerations, the effective moment of inertia (Ie) is calculated using the following formula: le 1 - 1cr ((2/3) Mcr) Ma 2 - وا ≥ Note that the upper bound was previously set as Iut in the earlier ACI equation. (a) Arrange the following moment of inertia values in ascending order (from smallest to largest): le, Ier, Ig and lut (b) Mer is the cracking moment. Choose the cross-section that should be used to compute Mcr. NA. h 5. Identify and circle the figure that represents the scenario in which the torsional effect is permitted to be reduced according to the ACI code provisions. (3 pts) mt mi B (b)arrow_forward
- I will rate, thanksarrow_forward. 9. A reinforced concrete beam is subjected to V/ = 40 kips and Tu/ = 12 ft kips at the critical section. Given conditions: ⚫ Longitudinal reinforcements use No. 8 grade 60 steel with an effective depth d = 20 in. For shear capacity, V = 18 kips and V₂ = 22 kips • For transverse reinforcements, use No. 3 bars with grade 60. • The effective torsional area of A. = 150 in². • Crack angle = 45° ⚫ The minimum stirrup spacing is Smin = 4" and the maximum stirrup spacing is Smax = Find the required stirrup spacing at the critical section. 8".arrow_forward3. The beam shown on the right uses three No. 8 bars made of Grade 60 steel as longitudinal reinforcement. The allowable maximum center-to-center spacing of the longitudinal rebars has been determined to be 10 inches. Now assume that Grade 80 steel will be used instead. Determine whether the beam satisfies the rebar spacing requirements according to the ACI Code. Additional assumptions: • Estimate fs = fy • 20" Clear cover: ? 12" Clear side cover: 1.5" The clear cover depth cc and the clear side cover remain unchanged, regardless of the change in material.arrow_forward
- Construction Materials, Methods and Techniques (M...Civil EngineeringISBN:9781305086272Author:William P. Spence, Eva KultermannPublisher:Cengage LearningEngineering Fundamentals: An Introduction to Engi...Civil EngineeringISBN:9781305084766Author:Saeed MoaveniPublisher:Cengage LearningMaterials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning
- Solid Waste EngineeringCivil EngineeringISBN:9781305635203Author:Worrell, William A.Publisher:Cengage Learning,Architectural Drafting and Design (MindTap Course...Civil EngineeringISBN:9781285165738Author:Alan Jefferis, David A. Madsen, David P. MadsenPublisher:Cengage LearningSteel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning





