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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
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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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
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
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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