The 200 x 200 x 1.250-mm oak (E-12 GPa) block (2) shown is reinforced by bolting two 5x 200 x 1.250 mm steel [E-200 GPa] plates (1) to opposite sides of the block. A concentrated load of 300 KN is applied to a rigid cap. Assume P-300 KN, L-1.25 m. Determine (a) the normal stresses in the steel plates (1) and the oak block (2). (b) the shortening of the block when the load is applied. A₁- Calculate the cross-sectional area of each steel plate. As, and the cross-sectional area of the oak block. Az Answers: A₂- (2) i Part 2 Save for Later (1) Answer: F₂/F- mm² mm² Attempts: 0 of 1 used Submit Answer Derive a compatibility equation using force-deformation relationships substituted into a geometry-of-deformation relationship. Express the compatibility equation here as the ratio of the force in the oak block, F₂, to the force in each steel plate. F
The 200 x 200 x 1.250-mm oak (E-12 GPa) block (2) shown is reinforced by bolting two 5x 200 x 1.250 mm steel [E-200 GPa] plates (1) to opposite sides of the block. A concentrated load of 300 KN is applied to a rigid cap. Assume P-300 KN, L-1.25 m. Determine (a) the normal stresses in the steel plates (1) and the oak block (2). (b) the shortening of the block when the load is applied. A₁- Calculate the cross-sectional area of each steel plate. As, and the cross-sectional area of the oak block. Az Answers: A₂- (2) i Part 2 Save for Later (1) Answer: F₂/F- mm² mm² Attempts: 0 of 1 used Submit Answer Derive a compatibility equation using force-deformation relationships substituted into a geometry-of-deformation relationship. Express the compatibility equation here as the ratio of the force in the oak block, F₂, to the force in each steel plate. F
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![The 200 x 200 x 1,250-mm oak [E-12 GPa] block (2) shown is reinforced by bolting two 5 x 200 x 1,250 mm steel [E - 200 GPa]
plates (1) to opposite sides of the block. A concentrated load of 300 kN is applied to a rigid cap. Assume P-300 kN, L-1.25 m.
Determine
(a) the normal stresses in the steel plates (1) and the oak block (2).
(b) the shortening of the block when the load is applied.
B
(1)
Answers:
A₁-
A₂-
i
Part 2
Calculate the cross-sectional area of each steel plate, A₁, and the cross-sectional area of the oak block, Az.
i
Save for Later
P
CL
(2) C
Answer: F₂/F₁-
CL
Save for Later
CL
(1)
mm²
mm²
Derive a compatibility equation using force-deformation relationships substituted into a geometry-of-deformation relationship.
Express the compatibility equation here as the ratio of the force in the oak block, F₂, to the force in each steel plate, F₁.
Attempts: 0 of 1 used Submit Answer
Attempts: 0 of 1 used Submit Answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec0472f3-d81c-4d4e-8d23-e88643cce0bb%2F47cc90ee-301c-49d4-9dca-1d81fcea4a4b%2Fzkng5u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The 200 x 200 x 1,250-mm oak [E-12 GPa] block (2) shown is reinforced by bolting two 5 x 200 x 1,250 mm steel [E - 200 GPa]
plates (1) to opposite sides of the block. A concentrated load of 300 kN is applied to a rigid cap. Assume P-300 kN, L-1.25 m.
Determine
(a) the normal stresses in the steel plates (1) and the oak block (2).
(b) the shortening of the block when the load is applied.
B
(1)
Answers:
A₁-
A₂-
i
Part 2
Calculate the cross-sectional area of each steel plate, A₁, and the cross-sectional area of the oak block, Az.
i
Save for Later
P
CL
(2) C
Answer: F₂/F₁-
CL
Save for Later
CL
(1)
mm²
mm²
Derive a compatibility equation using force-deformation relationships substituted into a geometry-of-deformation relationship.
Express the compatibility equation here as the ratio of the force in the oak block, F₂, to the force in each steel plate, F₁.
Attempts: 0 of 1 used Submit Answer
Attempts: 0 of 1 used Submit Answer
![Part 3
Consider a free body diagram of the rigid cap, and solve for the force in each steel plate, F₁, and the force in the oak block, F2.
based on the result from Part 2 and the applied load of 300kN. By convention, a tension force is positive, and a compression force
is negative.
Answers:
F₁-
F₂-
Save for Later
Part 4
Answers:
01 =
Calculate the normal stress in each steel plate, 0₁, and the normal stress in the oak block, 0₂. By convention, a tension stress is
positive, and a compression stress is negative.
0₂ =
i
Part 5
i
Save for Later
Answer: ₂ =
KN
Save for Later
kN
i
MPa
MPa
Calculate the shortening of the oak block. Since the oak block is shortened, this will be a negative value.
Attempts: 0 of 1 used Submit Answer
mm
Attempts: 0 of 1 used Submit Answer
Attempts: 0 of 1 used
Submit Answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec0472f3-d81c-4d4e-8d23-e88643cce0bb%2F47cc90ee-301c-49d4-9dca-1d81fcea4a4b%2F3pkv4t6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Part 3
Consider a free body diagram of the rigid cap, and solve for the force in each steel plate, F₁, and the force in the oak block, F2.
based on the result from Part 2 and the applied load of 300kN. By convention, a tension force is positive, and a compression force
is negative.
Answers:
F₁-
F₂-
Save for Later
Part 4
Answers:
01 =
Calculate the normal stress in each steel plate, 0₁, and the normal stress in the oak block, 0₂. By convention, a tension stress is
positive, and a compression stress is negative.
0₂ =
i
Part 5
i
Save for Later
Answer: ₂ =
KN
Save for Later
kN
i
MPa
MPa
Calculate the shortening of the oak block. Since the oak block is shortened, this will be a negative value.
Attempts: 0 of 1 used Submit Answer
mm
Attempts: 0 of 1 used Submit Answer
Attempts: 0 of 1 used
Submit Answer
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