Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Question
Chapter 4, Problem 4.8.2P
To determine
The maximum strength of the given section.
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
The lap joint in the figure is fastened by four 20 mm Ø A325X bolts. Calculate the maximum safe load P that can be applied if the plates are 25
mm thick A-36 steel Use ASD.
4 @ 40 mm
P
2 @ 60 mm
P
Compute the nominal compressive strength of the member shown in Figure . Use AISC Equation E3-2 or E3-3.
A 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
Chapter 4 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
Ch. 4 - Prob. 4.3.1PCh. 4 - Prob. 4.3.2PCh. 4 - Prob. 4.3.3PCh. 4 - Prob. 4.3.4PCh. 4 - Prob. 4.3.5PCh. 4 - Prob. 4.3.6PCh. 4 - Prob. 4.3.7PCh. 4 - Prob. 4.3.8PCh. 4 - Prob. 4.4.1PCh. 4 - Prob. 4.4.2P
Ch. 4 - Prob. 4.6.1PCh. 4 - Prob. 4.6.2PCh. 4 - Prob. 4.6.3PCh. 4 - Prob. 4.6.4PCh. 4 - Prob. 4.6.5PCh. 4 - Prob. 4.6.6PCh. 4 - Prob. 4.6.7PCh. 4 - Prob. 4.6.8PCh. 4 - Prob. 4.6.9PCh. 4 - Prob. 4.7.1PCh. 4 - Prob. 4.7.2PCh. 4 - Prob. 4.7.3PCh. 4 - Use A992 steel and select a W14 shape for an...Ch. 4 - Prob. 4.7.5PCh. 4 - Prob. 4.7.6PCh. 4 - Prob. 4.7.7PCh. 4 - The frame shown in Figure P4.7-8 is unbraced, and...Ch. 4 - Prob. 4.7.9PCh. 4 - Prob. 4.7.10PCh. 4 - Prob. 4.7.11PCh. 4 - Prob. 4.7.12PCh. 4 - Prob. 4.7.13PCh. 4 - Prob. 4.7.14PCh. 4 - Prob. 4.8.1PCh. 4 - Prob. 4.8.2PCh. 4 - Prob. 4.8.3PCh. 4 - Prob. 4.8.4PCh. 4 - Prob. 4.9.1PCh. 4 - Prob. 4.9.2PCh. 4 - Prob. 4.9.3PCh. 4 - Prob. 4.9.4PCh. 4 - Prob. 4.9.5PCh. 4 - Prob. 4.9.6PCh. 4 - Prob. 4.9.7PCh. 4 - Prob. 4.9.8PCh. 4 - Prob. 4.9.9PCh. 4 - Prob. 4.9.10PCh. 4 - Prob. 4.9.11PCh. 4 - Prob. 4.9.12P
Knowledge Booster
Similar questions
- 3. Determine the maximum tensile capacity of a L 178 x 102 x 19 angle as shown below if bolt diameter = 24 mm. The long leg of the member is connected to a 24mm thick guest plate. Use A572M gr.345 steel material. Ta 76 → 76 V L178x102x19 mam 76 76 5 38 4. The tension member shown below is a C 310 x 30.8 of A572M gr 345 steel materials. Will it safely support a service dead load of 265 kN and a service live load of 555 kN.arrow_forward4.3-4 Determine the available strength of the compression member shown in Figure P4.3-4. in each of the following ways: a. Use AISC Equation E3-2 or E3-3. Compute both the design strength for LRFD and the allowable strength for ASD. 15 HSS 10x6x ASTM A500, Grade B steel (Fy=46 ksi) 2/3arrow_forwardA 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. a) Determine the allowable tensile strength of the section based on tensile rupture of the net area.b) Determine the allowable tensile strength of the section based on yielding of the gross area.arrow_forward
- A PL 38 x 6 tension member is welded to a gusset plate as shown in figure. The steel is A36. PL ½ x 6 The design strength based on yielding is nearest to: The design strength based on rupture is nearest to: The design strength for LRFD is nearest to: The allowable strength based on yielding is nearest to: The allowable strenath based on rupture is nearest to: The allowable strength for ASD.arrow_forwardFor the clevis connection shown, determine the shear stress in the 23-mm-diameter bolt for an applied load of P = 165 kN. O 141 MPa O 211 MPa O 167 MPa O 120 MPa O 199 MPa Clevisarrow_forwardThe member shown in Figure has lateral support at points A, B, and C. Bending is about the strong axis. The loads are service loads, and the uniform load includes the weight of the member. A992 steel is used. Is the member adequate? a. Use LRFD. b. Use ASD.arrow_forward
- 7 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_forwardA PL 38 X 6 tension member is welded to a gusset plate as shown. The steel is A36 (Fy = 36ksi, Fu = 58ksi). a. The design strength, Pu based on gross area b. The design strength, Pu based on effective area PL % x 6 3/8" 6" Cross Sectional area of PL3/8x6 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answerarrow_forward2) Find the axial stresses of menbers FD, GD, GE State if it is tensile or Compressive. 4M 3M A LE 3m G 20 RN Go KNarrow_forward
- PROBLEM NO. 2: Two plates each with thickness t = 16mm are bolted together with 6-22mm diameter bolts forming a lap connection. Bolt Spacing are as follows: S₁ = 40mm, S₂ = 80mm, S3 = 100mm. Bolt diameter = 22mm. Use A36 steel. Calculate the allowable strength due to block shear only. I= 16 mm S₁ S₂ S₂ S1 40 80 80 40 18-16mm S-40 S₁-100 S-40arrow_forwardA W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by ¾ inch boltas shown below. Use A36 steel with Fy=36 ksi and Fu=58 ksi Determine the Yielding Capacity of the section based on LRFD (kips) Determine the Tensile Rupture capacity of the section based on LRFD Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFDarrow_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
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
Steel Design (Activate Learning with these NEW ti...
Civil Engineering
ISBN:9781337094740
Author:Segui, William T.
Publisher:Cengage Learning