(a)
The satisfaction of AISC interaction equation using LRFD.
Answer to Problem 6.2.1P
The member satisfies the AISC interaction equation.
Explanation of Solution
Given:
The load is
The length of member is
The value of
The flexural load is
Concept Used:
Write the LRFD interaction equation.
Here, the factored load is
Calculation:
Calculate the factored load.
Here, the dead load is
Substitute
Calculate the effective length of the member.
Here, the unsupported length is
Substitute
Calculate the axial compressive design strength.
From the manual table, the axial compressive design strength of a
Calculate the nominal flexural strength about x-axis.
From the design table, calculate the nominal flexural strength about x-axis by using
Calculate the flexural load about x-axis.
Here, the flexural dead load is
Substitute
There is no bending about y-axis, therefore
Write the equation to calculate the controlling interaction formula.
Substitute
The value is greater than
Calculate the LRFD interaction equation.
Substitute
The interaction equation is satisfied.
Conclusion:
Therefore, the interaction equation is satisfied with the AISD interaction equation.
(b)
The satisfaction of AISC interaction equation using ASD.
Answer to Problem 6.2.1P
The member satisfies the AISC interaction equation.
Explanation of Solution
Concept Used:
Write the ASD interaction equation.
Here, the factored load is
Calculation:
Calculate the factored load.
Here, the dead load is
Substitute
Calculate the allowed compressive strength.
From the manual table, the allowed compressive strength of a
Calculate the nominal flexural strength about x-axis.
From the design table, calculate the nominal flexural strength about x-axis by using
Calculate the flexural load about x-axis.
Here, the flexural dead load is
Substitute
There is no bending about y-axis, therefore
Write the equation to calculate the controlling interaction formula.
Substitute
The value is greater than
Calculate the ASD interaction equation.
Substitute
The interaction equation is satisfied.
Conclusion:
Therefore, the interaction equation is satisfied with the ASD interaction equation.
Want to see more full solutions like this?
Chapter 6 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
- The problem is about Tension Member for Steel Design. Please help me understand and answer it. I will leave a like. Thank you.arrow_forwardCheck the slenderness of the steel section against Local Buckling. 340 mm Use Fy = 800 MPa. O Compact Slender I O Adequate 400 mm None of the choices 30 mm 20 mm 20 mmarrow_forwardThe aluminum rod AB (Gal = 27 GPa) is bonded to the brass rod BD (Gbr = 39 GPa). The portion CD of the brass rod is hollow and has an inner diameter of 40 mm. If the allowable shear stresses for the aluminum and brass are 90 MPa and 70 MPa, respectively, and the allowable angle of twist of the free end A is 6 degrees, determine the maximum permissible value of T. Please show your complete handwritten solution. Thank youarrow_forward
- A flexural member is fabricated from two flange plates 1/2in x 16in and a web plate 1/4in x 20in. The yield stress of the steel is 50 ksi. a. Compute the elastic section modulus S and the yield moment My with respect to the major principal axis. b. Compute the plastic section modulus Z and the plastic moment Mp with respect to the major principal axis.arrow_forwardProblem 9.56 The C83400-red-brass rod AB and 2014-T6-aluminum rod BC are joined at the collar B and fixed connected at their ends. The cross-sectional area of each member is 1130 mm There is no load in the members when T= 10°C (Eigure 1) Figure 1 of 1arrow_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_forward
- TENSION MEMBERS: THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P. a. P BASED ON TENSION OF GROSS AREA b. P BASED ON TENSION OF NET AREA c. P BASED ON BEARING OF PROJECTED AREA d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)arrow_forwardThe member consists of the steel rod AB that is screwed into the end of the bronze rod BC. Find the largest value of P that meets the following design criteria: (i) the overall length of the member is not to change by more than 3 mm; and (ii) the stresses are not to exceed 140 MPa in steel and 120 MPa in bronze. The moduli of elasticity are 200 GPa for steel and 80 GPa for bronze. A Steel 1.0 m A = 480 mm² В 3P Bronze A = 650 mm² 2 m 2Parrow_forwardThe member consists of the steel rod AB that is screwed into the end of the bronze rod BC. Find the largest value of P that meets the following design criteria: (i) the overall length of the member is not to change by more than 3 mm; and (ii) the stresses are not to exceed 140 MPa in steel and 120 MPa in bronze. The moduli of elasticity are 200 GPa for steel and 80 GPa for bronze.arrow_forward
- 2- A 2m COPPER BAR SUBJECTEO TO A TENSILE IS SUSPENDED FROM A LOAD IBO KN SUPPORTED BY TWO IDENTICAL PIN THAT S pOSIS AS SHON OF THE LOWER AFPLIED TENSILE POSTS SUPPORT THE STEEL DETERMINE THE TOTAL DEFORMATION END OF THE LOAD HINT: NOTE THAT THE STEEL PIN WHERE THE COFPER BAR IS ATTACHED. COPPER BAR DUE TO STEEL POST L=0.5m A= 4500mm E- 200GPA COPFER BAR L- am A= 4800 Mn E- l20 GPa 180KNarrow_forwardH.W4: a- A copper bar is 900 mm long and circular in section. It consists of 200 mm long bar of 40 mm diameter, 500 mm long bar of 15 mm diameter and 200 mm long bar of 30 mm diameter. If the bar is subjected to a tensile load of 60 kN, find the total extension of the bar. Take E for the bar material as 100 GPa. b- A stepped bar ABCD consists of three parts AB, BC and CD such that AB is 300 mm long and 20 mm diameter, BC is 400 mm long and 30 mm diameter and CD is 200 mm long and 40 mm diameter. It was observed that the stepped bar undergoes a deformation of 0.42 mm, when it was subjected to a compressive load P. Find the value of P, if E = 200 GPa.arrow_forwardThe 2014-T6 aluminum rod AC is reinforced with the firmly bonded A992 steel tube BC. When no load is applied to the assembly, the gap between end C and the rigid support is 0.5 mm. Determine the support reactions when the axial force of 400 kN is applied. 400 mm 400 kN B A992 steel 50 mm 800 mm 25 mm 2014-T6 aluminum alloy Section a-a -400(10*) N st (a)arrow_forward
- Steel Design (Activate Learning with these NEW ti...Civil EngineeringISBN:9781337094740Author:Segui, William T.Publisher:Cengage Learning