SITUATION II: Calculate the permissible tensile load P 50 under the following conditions: Fy3248 MPa; Fu=600 MPa. The diameter of the bolts is 16 mm. V16 A 50 100 50 50 100 100 50 50 Allowable stresses: Tensile stresses on gross area: 0.6Fy Shear stress of the bolts: Fv=130 MPa Bearing stress of the bolts: Fp=1.2Fu 7. Based on shear capacity of bolts. a. 156.83 kN b. 1105.92 kN C. 475.21 kN d. 450.78 kN 8. Based on the bearing capacity of bolts. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 kN 9. Based on the yielding of plates. b. 1105.92 kN d. 450.78 kN a. 156.83 kN C. 475.21 kN

SITUATION II: Calculate the permissible tensile load P 50 under the following conditions: Fy3248 MPa; Fu=600 MPa. The diameter of the bolts is 16 mm. V16 A 50 100 50 50 100 100 50 50 Allowable stresses: Tensile stresses on gross area: 0.6Fy Shear stress of the bolts: Fv=130 MPa Bearing stress of the bolts: Fp=1.2Fu 7. Based on shear capacity of bolts. a. 156.83 kN b. 1105.92 kN C. 475.21 kN d. 450.78 kN 8. Based on the bearing capacity of bolts. a. 156.83 kN C. 475.21 kN b. 1105.92 kN d. 450.78 kN 9. Based on the yielding of plates. b. 1105.92 kN d. 450.78 kN a. 156.83 kN C. 475.21 kN

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: The 175 x 100 x 16 mm angle shown in the figure 3.28 connected to a gusset plate with three 25 mm Ø…

Q: A PL ½ x 8 tension member is connected with six 1-inch diameter bolts as shown in the figure below.…

A: Answer We are given Plate 1/2 x 8 bolt diameter = 1 in A36 steel Shear lag factor U = 0.1

Q: For the given riveted connection shown, having the allowable stresses: Fy = 248 MPa Fy = 150 MPa…

Q: Two plates are held together by 3, 23 mm diameter bolts. If a 11 kN load is applied to the plates in…

A: Number of bolt = 3 Diameter of bolt (d) = 23 mm Load (P) = 11KN

Q: For the two lines of M20 bolts shown in the figure, determine the design tensile strength (kN) of…

A: given- a bolted section. required- to find design tensile strength.

Q: 6.8 Problem: A bracket shown in figure 6.8 carries an eccentric load of 80 kN at an eccentricity o…

Q: A 3/4 in (thickness) X 5 in (height) steel plate is connected to another one through two bolts of…

Q: Material Strengths: Fy = 248 MPa, Fu = 400 M Pa For the two lines of bolt holes, the pitch is the…

Q: 0: The bracket shown carries an eccentric load P at an eccentricity e = 400 mm. the column and…

A: Given data:(a) Polar Moment of inertia

Q: W 10x45 is connected to two plates with two lines of ¾” bolts as shown below. Determine the tensile…

Q: 1-66 Figure P1-69 shows a bolt subjected to a tensile load. One failure mode would be, if the…

Q: The bolts in the lap joint shown are 22mm in diameter and 25mm diameter holes. The plates are A36…

Q: Determine the maximum factored LRFD tensile force capacity in tension only (no block shear). The…

A: We know that Maximum factored LRFD tensile force capacity of the member is given by minimum of the…

Q: A butt joint shown in figure 3.29 is bolted using A 325 bolts with a diameter of 22 mm, with threads…

Concept explainers

Bars

In civil engineering, bars refer to steel reinforcement bars or rebars. These are structural members primarily used for reinforcement purposes in masonry works. A bar is characterized by its cross-sectional area, which is negligible compared to its length. Bars are usually manufactured by steel which is uniform throughout the length, or by meshing multiple steel wires. Other materials of the bar include carbon steel, glass fiber, carbon fiber, and basalt fiber. The reinforcing bars are coated with epoxy resins and other materials to enhance some of their properties including corrosion resistance.

Question

100%

SITUATION II:
Calculate the permissible tensile load P
350
under the following conditions: Fy-248
MPa; Fu=600 MPa. The diameter of the
bolts is 16 mm.
716
100
50
50
100
100
50
50
Allowable stresses:
Tensile stresses on gross area: 0.6Fy
Shear stress of the bolts: Fv=130MPA
Bearing stress of the bolts: Fp=1.2Fu
7. Based on shear capacity of bolts.
a. 156.83 kN
C. 475.21 kN
b. 1105.92 kN
d. 450.78 k
8. Based on the bearing capacity of
bolts.
a. 156.83 kN
C. 475.21 kN
b. 1105.92 kN
d. 450.78 kN
9. Based on the yielding of plates.
a. 156.83 kN
C. 475.21 kN
b. 1105.92 kN
d. 450.78 kN

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Step 5

Step by step

Solved in 5 steps with 5 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Subject: Steel design- Bolted Steel Connection - Bolted Subjected to Eccentric Loads *Use NSCP 2015 formula/guide to solve this problem *Use Handwritten
How
Use ASD to determine the width of the 5/8 inch thick A36 plate subjected to tension as shown below. The dead load effect is PD=85 kip while the live load effect is PL=80 kip. Two rows of 5/8 inch standard bolts will be used for the connection and the plate width can be manufactured in 1 inch increments. Assume that the larger connected plate does not fail. Consider the limit states of gross tensile yield and tensile rupture. Find the required tension strength, Pa, for the controlling ASD load combination. Determine the width necessary for the plate to resist gross tensile yield. Determine the width necessary for the plate to resist tensile rupture. Report the required plate width.
2. For the figure shown, calculate the net area of the 10 x 200 mm plate. The plate is connected at its end with two lines of 20 mm bolts. Using LRFD, determine the load P that the plates can carry against tensile yielding in the gross section and the tensile rupture in the net section. Use A-36 steel. PL 5x200mm PL 5x200mm Pu PL 10x200mm Pu
Please solve
Calculate all 5 failures paths. Show step by step solution. A flat bar PL 12.7mm X 165mm Cross sectional area is 2095.5mm2. All units to be converted from SI to metric. Textbook reference: Handbook of steel construction CISC
.
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
Can u solve me step by step please . thank u
Please solve the following problem is ksi
Please include the detailed explanation and diagram
SITUATION I A lapped bolted lension member is shown. Diameter of bolts are 22 mm and the plate material is A36 steel Fy = 275 MPа, Fи 3 415 МPа. Assume the fasteners are adequate and do not control the tensile capacity. Diameter of hole is 2.5 mm bigger than the diameter of bolt. 12 :: 350 50 100 100 50 50 1. Determine the tensile capacity of the lapped joint based on gross area. a. 252.656 kN c. 866.250 kN b. 202.125 kN d. 693.000 kN 2. Determine the net Area for the tensile capacity of the lapped joint. а. 3318 тm? c. 5250 mm? b. 1554 mm? d. 1824 mm? 3. Determine the tensile capacity of the lapped joint based on the net area. a. 752.856 kN C. 866.250 kN b. 688.845 kN d. 1039.500 kN