(a)
Whether a
Answer to Problem 5.5.15P
Inadequate
Explanation of Solution
Given:
A992 steel
Formula used:
Calculation:
Determine the nominal flexural strength:
From the
For
This shape is compact because there is no footnote in the dimensions and properties tables to indicate otherwise.
Since
Determine the factored point loads:
D is dead point load and L is live point load
Determine the factored distributed loads:
Determine the beam reactions:
Compute Cb :
Since
Conclusion:
(b)
Whether a
Answer to Problem 5.5.15P
Inadequate
Explanation of Solution
Given:
A992 steel
Formula used:
Calculation:
Determine the nominal flexural strength:
From the Zxtable,
For
This shape is compact because there is no footnote in the dimensions and properties tables to indicate otherwise.
Since
Determine the factored point loads:
Determine the factored distributed loads:
Determine the beam reactions:
Compute Cb :
Since
Conclusion:
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Chapter 5 Solutions
STEEL DESIGN W/ ACCESS
- Problem 4: Use A992 steel and select a W shape. The service dead load is 142 kips, and the service live load is 356 kips. ↑ a 10' wwww Do wwwwww y-axis 16' x-axisarrow_forwardA column is built up from four (4)- 125 x 125 x 18 angle shapes as shown. The plates are not continuous but are spaced at intervals along the column length and function to maintain the separation of the angles. They do not contribute to the cross-sectional properties. The effective length is 4 m. Compute the allowable design compressive strength based on flexural buckling. E= 250 MPa. Use ASD. k 375 mm 125mm, HPlate 125mm 4 - 4 125 × 125× l8 section 下好业arrow_forwardEstimate the cross-sectional area of a 350S125-27 cold-formed shape. a. If the member is tested in tension, what would be the maximum force thesample could carry before reaching the yield strength if the steel has ayield strength of 225 MPa?b. Would you expect a 2.5 m stud to carry the same load in compression?(explain)arrow_forward
- Steel design Use A992 steel and select a W shape. The service dead load is 142 kips, and the service live load is 356 kips.arrow_forwardDetermine the maximum axial compressive service load that can be supported if the live load is twice as large as the dead load. Use AISC Equation E3-2 or E3-3. a. Use LRFD. b. Use ASDarrow_forwardA built-up section shown, 30 feet long, is fixed at both ends and braced in the weak direction at the midheight. A992 steel is used. Determine the following: 11. Ix 12. Ky 13. effective slenderness ratio 14. Fe 15. LRFD design strength 16. ASD allowable strength 18 4 W10 × 49 A 14.4 in² bf 10 in d. 10 in tw 0.340 in tf 0.560 in Ix= 272 in¹ Iy=93 4in¹ W10×54 A 15-3 in² bf 10 in d = 10 lin tw0-370 in tf 0.615 in Ix= 303 in 4 Ty = 103 in 1arrow_forward
- Q2) The members of the truss structure shown below is plain concrete. The compressive strength of the concrete is 25 MPa. Compute the maximum load P that can be carried by the structure. (Cross section of each member of the truss is 200 x 200 mm and don't use material factors and do not consider slenderness) Comment on your results briefly. P A& 2m SC 2 m 1380 2m Darrow_forwarddetermine the member stresses of the truss shown. indicate if the stress in the member is in tension or compression. present complete and neat solutions. p1-30kN W-25kN/m P2=51kN AH Av Im P₂ Im P₂ Im P₂ Im Av W 1.6marrow_forwardDesign the reinforcements of the given T beam below. bf=800mm bw=450mm tf=120mm d=600mm d'=80mm fc'=35MPa fy=350MPa USE NSCP 2015 A.Mu = 1300kN-m, As = B.Mu = 1600kN-m, As = mm2 _mm2, As' = mm2arrow_forward
- A simply supported beam is subjected to a uniform service dead load of 1.0 kips/ft (including the weight of the beam) and uniform service live load of 2.5 kips/ft. The beam is 40 feet long, The beam has continuous lateral support, and A572 Grade 50 steel is used. A572 Grade 50 steel has Fy = 50 ksi and Fu = 65 ksi. Is a W30 × 116 adequate from flexure and shear?Check both LRFD and ASD. Is the beam adequate from deflection if it allowsmaximum deflection of L/360 (dead load + liveload)?arrow_forward7 7a 7b 7c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is structural tubing however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? The section is 8" x 8" x 7.94 mm thick: Use Fy=248 MPa: E=200,000 MPa AISC wall thickness Ix 106 S x 103 Jx 103 mm4 mm3 mm4 rx =ry Area Ag (mm2) mm Designation Weight/m 8x8 7.94 47.36 6,039 79.25 37.84 371.99 60.35 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) 8x8 14.29 80.61 10,258 76.2 59.52 585.02 99.06 8x8 72.7 9,290 76.96 54.53 539.13 90.32 8x8 9.53 56.09 7,161 78.48 44.12 432.62 70.76 mm 12.7 Zx 103 mm3 437.53 714.48 650.57 512.92arrow_forwardThe beam shown in the figure has continuous lateral support of bothflanges. The uniform load is a service load consisting of 50% deadload and 50% live load. The dead load includes the weight of thebeam. If A992 steel is used, is a W16 × 31 adequate?a. Use LRFD.b. Use ASD.arrow_forward
- Steel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning