1. In the butt joint shown below, the 7/8-inch diameter bolts are made of A325 steel with threadsexcluded from the shear plane and the plates are made of A36 steel.12 in. xin.10012 in.a) Determine the bolt shear strengthb) Determine the bearing strength of the platec) Determine the tensile strength (both gross and net) of the plated) What is the strength of the bolted connection?

The width of the plate is B=12 in. The thickness of the plate is t=58 in. The diameter of the bolt…

Answered: 1. In the butt joint shown below, the…

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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The tension member shown in the figure below is a ¹/2 x 10 plate of A36 steel: Fy = 36 ksi, Fu = 58 ksi. The connection is with 9/8-inch-diameter bolts. -→| 20 |einetein | 21 | 6 spaces @2" t = 1/2 in. Compute the nominal strength based on the net section. (Express your answer to three significant figures.) P₁ = kips
For the cross section shown, calculate the tensile capacity of the steel plate. Assume that the bolt hole is 1/16 inch larger than bolt diameter. Plate 1/2" x 11" .T 7/8" dia. bolt, typ. ASTM A325N
A double lap splice connection shown in figure B is made up of 6mm x 150mm and 9mm x 150mm. Use NSCP specifications and assume a double row of bolts with standardsize holes for a bearing type connection with threads excluded from the shear planes.Minimum tensile strength Fu = 552.93MPa. Assume diameter hole is 3mm bigger than thediameter of bolts. Allowable shear stress is Fv = 207MPa. Allowable bearing stress fb = 1.2Fu.Diameter of bolts is 19mm. Fy = 449.25MPa.(a) Compute the tensile capacity of the steel plates. (Max. Net Area = 0.85 of Gross Area)(b) Compute the number of 19mm diameter bolts to develop the full strength of the plates. Note: SHOW DRAWINGS AND SOLUTIONS.
Two steel plate tension members have been connected using 0.72 inch diameter bolts arranged in an equally-spaced 9 by 3 rectangular formation. Both plates have a thickness equal to 1/3 in. Take shear fracture stress as 51.2 ksi, clear distance for each edge bolt as 0.12 in, and the clear distance for the other bolts as 1.072 in. a)Sketch the connection showing all the details and measurements with units. b)Find the maximum allowable service dead load and live load assuming live load is half as much as dead load. Consider ASD.
A plate with width of 420 mm and thickness of 13 mm is to be connected to a plate of the same width and thickness by 30 mm diameter bolts, as shown in the Figure 1. The holes are 3mm larger than the bolt diameter. The plate is A36 steel with yield strength Fy = 248MPa. Assume allowable tensile stress on net area is 0.60Fy. It is required to determine the value of b such that the net width along bolts 1-2-3-4 is equl to the net width along bolts 1-2-4. W = 420 mm t = 13 mm a = 64 mm c = 104 mm d = 195 mm Bolt Diameter = 30 mm Holes Diameter = 30 mm + 3 = 33 mm ?? = 248 MPa Allowable Tensile Stress on ?? = 0.60Fy a. Calculate the vaue of b in millimeters. b. Calculate the value of the net area for tension in plates in square millimeters. c. Calculate the value of P so that the allowable tensile stress on net area will not be exceeded.
1. In the butt joint shown below, the 7/8-inch diameter boits are made of A325 steel with threads excluded from the shear piane and the plates are made of A36 steel. o of a Determine the bolt shear strength b) Determine the bearing strength of the plate ) Determine the tensile strength (both gross and net) of the plate d) What is the strength of the bolted connection?
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…
7. Determine the design tensile strength of the plate (200 x 10 mm) connected to 12 mm thick gusset, using 24 mm bolts as shown below. Use A572M Grade 345 steel materials. 200 40 50 40 30 30 30 30 40
Problem (4): A36 steel is used for the tension member shown in Figure below. a. Determine the nominal strength based on the gross area. b. Determine the nominal strength based on the net area. PL 300mmx17mm 150 75 75 75 75 75 125 ImmmmTmm"" mmn mm mm 50mm 100mm 100mm 50mm 19mm diameter bolts Problem (5): The tension m shown in Figur
A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An. Section 2L5 x 3 x16 LLBB Determine the allowable tensile strength for ASD. Select one: а. 52.1 b. 104.2 c. 87.8 d. 43.9

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