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A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mm
- a)38
- b)39
Correct answer is between ' 38, 39'. Can you explain this answer?
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A plate 50 mm wide and 10 mm thick is to be welded as T joint to anot...
Thickness of the plate to be joined
h = 10 mm,τallow = 144 MPa
l = 38.3 mm
Most Upvoted Answer
A plate 50 mm wide and 10 mm thick is to be welded as T joint to anot...
Problem Statement:
A plate with dimensions 50 mm wide and 10 mm thick is to be welded as a T joint to another plate using two parallel fillet welds. The plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. We need to determine the weld length per side.
Given:
Width of the plate (b) = 50 mm
Thickness of the plate (t) = 10 mm
Load (F) = 78 kN
Allowable shear stress (τ) = 144 MPa
Assumptions:
1. The weld is assumed to have a triangular shape.
2. The fillet weld has a constant leg size.
3. The weld material has the same strength as the base material.
Solution:
Step 1: Calculate the shear force on each weld:
The total shear force on the weld is equal to the load applied to the joint. Since there are two parallel fillet welds, we divide the load equally between them.
Shear force on each weld = F/2 = 78 kN / 2 = 39 kN
Step 2: Calculate the area of each fillet weld:
The area of a triangular weld can be calculated using the formula:
Area of triangular weld = 0.707 * leg size * length of weld
We need to find the length of the weld. Let's assume the length of the weld per side is L.
Area of each fillet weld = 0.707 * leg size * L
Step 3: Calculate the shear stress in the weld:
The shear stress in the weld can be calculated using the formula:
Shear stress = Shear force / Area of weld
Substituting the values, we get:
Shear stress = 39 kN / (0.707 * leg size * L)
Since the allowable shear stress is given as 144 MPa, we can write:
144 MPa = 39 kN / (0.707 * leg size * L)
Step 4: Solve for the length of the weld:
Rearranging the equation, we get:
L = 39 kN / (0.707 * leg size * 144 MPa)
Converting the units to mm and MPa:
L = 39000 N / (0.707 * leg size * 144 N/mm^2)
Simplifying the equation, we get:
L = 39000 / (0.707 * leg size * 144)
L = 38.83 / leg size
Since the length of the weld must be a whole number, the length is between 38 and 39 mm.
A plate with dimensions 50 mm wide and 10 mm thick is to be welded as a T joint to another plate using two parallel fillet welds. The plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. We need to determine the weld length per side.
Given:
Width of the plate (b) = 50 mm
Thickness of the plate (t) = 10 mm
Load (F) = 78 kN
Allowable shear stress (τ) = 144 MPa
Assumptions:
1. The weld is assumed to have a triangular shape.
2. The fillet weld has a constant leg size.
3. The weld material has the same strength as the base material.
Solution:
Step 1: Calculate the shear force on each weld:
The total shear force on the weld is equal to the load applied to the joint. Since there are two parallel fillet welds, we divide the load equally between them.
Shear force on each weld = F/2 = 78 kN / 2 = 39 kN
Step 2: Calculate the area of each fillet weld:
The area of a triangular weld can be calculated using the formula:
Area of triangular weld = 0.707 * leg size * length of weld
We need to find the length of the weld. Let's assume the length of the weld per side is L.
Area of each fillet weld = 0.707 * leg size * L
Step 3: Calculate the shear stress in the weld:
The shear stress in the weld can be calculated using the formula:
Shear stress = Shear force / Area of weld
Substituting the values, we get:
Shear stress = 39 kN / (0.707 * leg size * L)
Since the allowable shear stress is given as 144 MPa, we can write:
144 MPa = 39 kN / (0.707 * leg size * L)
Step 4: Solve for the length of the weld:
Rearranging the equation, we get:
L = 39 kN / (0.707 * leg size * 144 MPa)
Converting the units to mm and MPa:
L = 39000 N / (0.707 * leg size * 144 N/mm^2)
Simplifying the equation, we get:
L = 39000 / (0.707 * leg size * 144)
L = 38.83 / leg size
Since the length of the weld must be a whole number, the length is between 38 and 39 mm.
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A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer?
Question Description
A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer?.
A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer?.
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A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer?, a detailed solution for A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer? has been provided alongside types of A plate 50 mm wide and 10 mm thick is to be welded as T joint to another plate by means of two parallel fillet welds. If the plates are subjected to a static load of 78 kN and the allowable shear stress is 144 MPa. The weld length per side (in mma) 38b) 39Correct answer is between ' 38, 39'. Can you explain this answer? theory, EduRev gives you an
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