5. Analysis of Tension MembersCompute the net area of an L6 x 3 ½ x 3/8 with two lines of 3/4"diameter bolts in the long leg and one line of 3/4" diameter bolts in theshort leg.✓ 16×3/2 × 3/8 (A = 3.44 IN²)Bolts are 3/4"L

Step 1: Calculate the Gross Area of the Angle Given dimensions are L6×321×83, where: 6 inches is…

Answered: 5. Analysis of Tension Members Compute…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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How many M24 bolts Grade 8.8 will be needed in a tension splice connected to a 20 mm cover plates on the longer leg of a 200 x 100 x 15mm angle in S355 steel, if the full strength of the angle is to be developed?
Compute the net area of each given members. 1. PLx 8 in Ø bolts 2. in O bolts W12 x 40- -PL를 x 14 3. WT15 x 45
Which of the following gives the number of M24 A325 bolts needed that will carry prying forces with factored load = 720 KN. O 6 O 3 O 4 O 5
In mm) Problem 8: A C310 × 45 is connected with 25mm diameter bolts in each flange, as shown in Figure, Problem 11. If Fy = 345MPA, Fu = 450MPA, and Ae = 0.90An, Apply the following %3D %3D %3D approaches to determine: a. The design strength for LRFD. b. The allowable strength for ASD. Section Figure. Problem 8
1. The tension member shown in Figure 3.4-2 is a PL 58 x 10, and the steel is A36. The bolts are 3/4-inch in diameter. a. Draw the different potential failure lines. b. Compute the effective net area. 22" 2½" 2" 4 3" *₁ 3" ✓ 2" 2. A double-channel shape, 2C10 x 20 steel, Fy = 50 ksi and Fu = 70 ksi, is used for a built-up tension member as shown in the figure. The holes are for 12-inch-diameter bolts. PNA a. Determine the design strength for LRFD. b. Determine the allowable strength for ASD. YD Shape A 2" 2" -X Area, Depth, d | 3²²-|-1 3²² | 00 "There is no elevator to success, you have to take the stairs." - Zig Ziglar God bless. Ⓒ O O O Web Thickness, tw in. in. O 4" C10-30 x25 x20 5.87 10.0 10 0.379 % x15.3 4.48 10.0 10 0.240 4 4" Table 1-5 C-Shapes Dimensions Flange Width, b₁ 2 in. in. in.2 in. C15x50 14.7 15.0 15 0.716 1168 3.72 3% 0.650 % x40 11.8 15.0 15 0.520 x33.9 10.0 15.0 15 0.400 % 14 3.52 32 0.650 % %/16 3.40 3% 0.650 % C12-30 8.81 12.0 12 0.510 4 3.17 3% x25 7.34 12.0 12…
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4. Given a sheared flat bar 5" in width, which is to have a double line of 3/4" Dia bolts, answer the following. a. Does the bolt hole pattern need to be staggered? b. What is the minimum edge distance required? c. What is the gage distance? d. What is the clearance "E" required? e. What is the diagonal minimum distance? f. What is the minimum pitch?
12. determine the capacity of the details shown below.A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksil) what is the bolt bearing capacity of beam flange of the flange connection in kips? (2 decimal places)
For high strength bolts with M20, 10.9 grade, please explain the meaning of '20', '10' and '.9'.
Situation 4: A steel plate is 300 mm wide and 15 mm thick with 4 bolts (24mm dia.) cut into the plate as shown. Hole diameter = diameter of bolt + 2 mm. Use Fy = 248 MPa and Fu = 400 MPa Given: S1 = 60mm, S2 = 55mm, S3 = 75mm, g1 = 80mm, g2 = 70mm, g3 = 70 mm $1 S2 T+ H g3 DO E $3 1. Which of the following nearly gives the Allowable Strength Ta? O 652 KN O 669 KN O 674 KN O 681 KN BO CO
(1) Develop a table showing the design strength. for A325-N, A325-x A490-N and 7 A 490-X bolts for the following sizes: "/2", 5/8", 3/4", 7/8", 1" x 1/8". Need to show steps for ½/2" for the table values. N X Size A325-N A490-N A325-X A490-X 1/₂ 5/8 314 7/8 1 1½8
Determine the capacity of the details shown below.A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksi l) what is the bolt bearing capacity of beam flange of the flange connection in kips? (2 decimal places)

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5. Analysis of Tension Members Compute the net area of an L6 x 3 ½ x 3/8 with two lines of 3/4" diameter bolts in the long leg and one line of 3/4" diameter bolts in the short leg. ✓ 16×3/2 × 3/8 (A = 3.44 IN²) Bolts are 3/4" L