EBK MATERIALS FOR CIVIL AND CONSTRUCTIO
4th Edition
ISBN: 8220102719569
Author: ZANIEWSKI
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 4, Problem 4.2QP
To determine
Complete the table for a member where deflection controls, which material require a larger cross section, and for a member where tension controls.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Complete Table P4.2. For a member where a deflection criterion controls,
which material would require a larger cross section? What about for a member
where tension controls?
TABLE P4.2
Mild Steel
7178 T76 Aluminum
Yield Strength
Ultimate Strength
Modulus of Elasticity
6. Give the properties table below.
A992 Steel
7178 T76 Aluminum
Yield Strength
50 ksi
73 ksi
Ultimate strength
75 ksi
85 ksi
Modulus of Elasticity
29,000 ksi
10,500 ksi
Assume you are designing a rectangular member with a specific allowed height.
For a design where deflection criterion controls which material would require a larger cross
section?
For a design where tension stress controls, which material requires the larger cross section?
} 8) A tensile load of 3000 N is applied to a steel bar of cross-sectional area 500 mm?. If
a tensile load was applied to an aluminum bar to achieve the same lateral (transvers) strain
with the former (steel bar), what would be the load?
Note: Original dimensions of the two bars are the same
(Est = 2.1 x 105 MPa, Eat = 0.703 × 105 MPa, vg = 0.3, vat = 0.33) E = º , v = - Saterat
Eaxial
Fatuminum™
3000 N
A-500 mm²
A=500 mm²
Fatuminum=?
3000 N
Steel
Aluminum
Chapter 4 Solutions
EBK MATERIALS FOR CIVIL AND CONSTRUCTIO
Ch. 4 - Name the two primary factors that make aluminum an...Ch. 4 - Prob. 4.2QPCh. 4 - An aluminum alloy specimen with a radius of 0.28...Ch. 4 - An aluminum alloy bar with a radius of 7 mm was...Ch. 4 - Decode the characteristics of a 6063 T831...Ch. 4 - A round aluminum alloy bar with a 0.6 in. diameter...Ch. 4 - An aluminum alloy bar with a rectangular cross...Ch. 4 - A round aluminum alloy bar with a 0.25-in....Ch. 4 - An aluminum alloy rod has a circular cross section...Ch. 4 - An aluminum alloy cylinder with a diameter of 3...
Ch. 4 - A 3003-H14 aluminum alloy rod with 0.5 in....Ch. 4 - The stressstrain relation of an aluminum alloy bar...Ch. 4 - An aluminum specimen originally 300 mm long is...Ch. 4 - A tension stress of 40 ksi was applied on a 12-in....Ch. 4 - A tension test was performed on an aluminum alloy...Ch. 4 - In Problem 4.15, plot the stressstrain...Ch. 4 - Referring to Figure 4.5, determine approximate...Ch. 4 - Prob. 4.18QPCh. 4 - A tensile stress is applied along the long axis of...Ch. 4 - A cylindrical aluminum alloy rod with a 0.5 in....Ch. 4 - Prob. 4.21QPCh. 4 - Discuss galvanic corrosion of aluminum. How can...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular cross section of 0.2011in.2 in area and a gage length (the length over which the elongation is measured) of 2.000 inches. The specimen was not loaded to failure. a. Generate a table of stress and strain values. b. Plot these values and draw a best-fit line to obtain a stress-strain curve. c. Determine the modulus of elasticity from the slope of the linear portion of the curve. d. Estimate the value of the proportional limit. e. Use the 0.2 offset method to determine the yield stress.arrow_forwardProblem #3: A wide-flange beam, constructed with A572 steel, is cooled to -30°C and is subjected to dynamic loading. There is an edge crack in the top flange of 16 mm, as demonstrated in the figure below. Hints: ● 7.5 mm ● crack a 13.1 mm X 203 mm Cross-Section 303 mm a = 16 mm -203 mm- Sg Determine the bending moment about the x-axis that will cause brittle fracture of the beam. Note that negative bending causes the top flange to be subjected to tensile stresses. Top View of Flange 13.1 mm As illustrated in the figure above, treat the top flange as an edge-cracked tension member. Use the simplified expression for computing K, but still verify that it applies. The average stress in the top flange can be determined by an My/I calculation, as opposed to a P/A calculation from previous problems. Compute y based on the mid-depth of the top flange. Ignore the "fillets" (i.e., the rounded portions of the I-section) when calculating the moment of inertia, I.arrow_forwardQ5 A specimen of magnesium having a rectangular cross section of dimensions (3.2 mm X 19.1 mm) and gage length 63.5 mm is deformed in tension. Using the load-elongation data tabulated as follows load (N) and Extension (mm), Find the standard mechanical properties. load 1380 2780 5630 7430 8140 9870 12850 | 14100 14340 13830 12500 Ext. 0.03 0.06 0.12 0.2 0.25 0.64 1.91 3.18 4.45 5.72 6.99 a) Draw the stress strain curve. b) Find the young modulus. c) Find the proof stress d) Find the ultimate tensile strength. e) Find the fracture stress f) Find the maximum strain.arrow_forward
- An alloy steel having a diameter of 0.45 in. extends 0.002 in. over a gage length of 2.2 in. due to a tensile load of 3900 lb. What is the modulus of elasticity of this alloy steel? A)29,010 ksi B)27,000 ksi C)30,000 ksi D)28,000 ksiarrow_forwardWhat is the value of reduction factor of the beam, if the stress in the steel is 400 MPa, and the Modulus of Elasticity of the Steel is 200,000 MPa. O 0.671 0.692 O 0.660 O 0.650 None of the Choices O 0.681arrow_forwardREINFORCEB=(335mm)H=(435mm)arrow_forward
- A force of 100,000 N is applied to an iron bar with a cross-sectional area of 10 mm × 20mm and having a yield strength of 400 MPa and a tensile strength of 480 MPa,Determinea. whether the bar will plastically deform; andb. whether the bar will experience neckingarrow_forward2- the rubber blocks shown in Figure 2 are used in a double U shear mount component. The dimensions of the rubber blocks are 15 x 21 x 25 mm and the applied load of 650 N cause the upper frame to be deflected by 5 mm. ↓ Double U anti-vibration shear mount 15 Shear deformation of blocks 25 Rubber block dimensions 21 오 Figure 2 (a) determine the average shear stress and shear strain in each of rubber blocks (b) determine the shear modulus G of the rubber material used in this component.arrow_forwardThe beam is made from a material that can safely support a tensile or compressive stress of 350 MPa. a) Based on this criterion, determine the largest force, F, the beam will safely support if it has the cross- section (a). b) What is the largest allowable F for cross-section (b) Recall for a rectangle: I = bh³/12 1.0 m- 60 mm 23.3 mm Figure 1 Cross-section (a) -0.63- 10 mm 10 mm 40 mm 10 mm -50 mm - Figure 2 Cross-section (b)arrow_forward
- Find the lightest S-shape for the beam shown if the working stress in bending is 175 MPa. What is the maximum bending stress in the beam selected? 50 kN/m 4 m -2.4 m· -x X- TABLE B-3 Properties of I-Beam Sections (S-Shapes): SI Units Flange Axis Y-Y Web X-X spxy Mass Area Width ıdə (wu) Thickness thickness Pl1 = S (10³ mm³) yli^ = 4 (10° mm*) (u) 27.7 27.7 22.1 Designation (10° mm*) (ww) 20.3 (ww) 38.9 („ww) (ww) (ww) 081 158 006 77 622 622 204 1 320 1 220 081 x 019S 4 230 240 34.5 338 × 158 x 149 x 134 007 184 15.7 3930 3 260 247 O'ZE 19.7 320 001 07 6'6E 32.3 006 81 00 I 15 200 149 6'81 15.9 166 937 677 234 019 215 134 181 22.1 090 € 2870 019 18.6 205 33.0 611 22.1 12.7 874 241 17.5 197 61I x 34.0 019 S510 × 143 143 18 200 16300 14 200 12 500 91S 91S 183 23.4 23.4 20.2 20.3 16.8 S69 653 2 700 20.8 228 x 128 x 112 × 98.2 128 112 33.8 34.5 29.5 179 961 2 540 007 194 19.4 80s 508 216 152 162 16.1 533 001 7 1950 12.3 98.2 20.2 12.8 495 6SI 11.4 144 30.2 661 S460 × 104 104 13 200 457 6SI…arrow_forwardA composite column member is fabricated on top of a base plate as shown in the figure. What should be the minimum safe thickness of the plate against punching if the total deformation of the column is 0.72 mm. Given below are the material properties. Material Oallowable (MPa) Tallowable (MPa) E (GPa) Steel 120 60 200 Aluminum 90 45 70 Steel (D = 55 mm) Aluminum (D = 110 mm) Steel 500 mm 300 mmarrow_forward2.) Answer the following: a.) a 22 ft long solid steel rod is subjected to a load of 15,000 lb. This load causes the rod to stretch 0.435 in. The modulus of elasticity of the steel is 30,000,000 psi. Determine the diameter of the rod. b.) A solid circular titanium control rod must support an axial tension force of 12,000 lb. Assume that the modulus of elasticity of the titanium in 16,500,000 psi. If the rod must elongate no further than 0.35 in. when the normal stress of the rod is 68,000 psi, determine the minimum rod diameter AND the maximum rod length.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Steel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning
Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning
Steel Design (Activate Learning with these NEW ti...
Civil Engineering
ISBN:9781337094740
Author:Segui, William T.
Publisher:Cengage Learning
Materials Science And Engineering Properties
Civil Engineering
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning