3. Determine the LRFD design tensile strength and the ASD allowable tensile strength for an A36 (Fy= = 36 ksi and Fu = 58 ksi) L6 x 6 x 3/8 inches that is connected at its ends with one line of four 7/8-in-diameter bolts in standard holes 3 inches on center in one leg of the angle. Note: L6 x 6 x 3/8 (Ag = 4.38 in²; y = x=1.62 in)

3. Determine the LRFD design tensile strength and the ASD allowable tensile strength for an A36 (Fy= = 36 ksi and Fu = 58 ksi) L6 x 6 x 3/8 inches that is connected at its ends with one line of four 7/8-in-diameter bolts in standard holes 3 inches on center in one leg of the angle. Note: L6 x 6 x 3/8 (Ag = 4.38 in²; y = x=1.62 in)

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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| * A's O As,web As bw Steel areas: A's-2 + 14 = 308 mm² A₁=50 mm² (area of $8) Materials : C25/30 / B420C Longitudinal reinforcements A's=2 $ 14 As,web= 2 $ 12 As= 5 0 14 bw=300 mm d=670 mm d'=30 mm k₁= 0,85 k3 = 0,85 Pb=0.0205 (from table) Transverse Reinforcements $8/200 mm (2-leg stirrups) Material coefficients Ymc:1.5 (Concrete) Yms: 1,15 (Steel) Determine the reinforcement ratio and failure type of the beam As,web= 212= 226 mm² Asw=2x50=100 mm² (Total Area of two leg stirrup) A₁= 5 + 14=770 mm²
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PL 5/8 x 12 3/4-in-diameter bolts
Determine the allowable strength of the figure as shown
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
PLEASE DOUBLE CHECK YOUR ANSWER Determine the maximum factored load that can be applied to the Tension member shown. The angle is ASTM A36. The diameter of the bolts is 1-in. CIVIL ENGINEERING STEEL DESIGN W12 x 16 4,-0.220 in -3.99 in 1=+=+=+= 3 in 3 in 3 in
Q14 What is the available tensile strength of a WT 20x74.5 made of A572 grade 50 steel? Use ASD, assume a shear lag factor (U) of 0.75, the connections are welded, and neglect block shear or other Chapter J requirements. Check both yielding and rupture. a) 800 kips b) 986 kips c) 533kips d) 656 kips
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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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