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Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer?.

Solutions for Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE). Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free.

Here you can find the meaning of Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Δ = displacement caused when force is increased by a small amountP = external force appliedN = internal force in the member force appliedL = length of memberA = cross-sectional area of memberE = Modulus of elasticitySame symbol is used for partial and total differentiation and they are pretty obvious.What will be change in work done in both case on initial application of load?a)p1dΔ1 + dp1 Δ1 + p2dΔ2b)1⁄2p1dΔ1 + dp1 Δ1 + p2dΔ2c)1⁄2p1dΔ1 +1⁄2dp1 Δ1 + p2dΔ2d)1⁄2p1dΔ1 +1⁄2dp1 Δ1 +1⁄2p2dΔ2Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.