-1.5 m 1'm -1.5 m. 0.75 m Probs. 8–51/52

Q: 400 N y 250 N B 60° Answers: 530 N 60⁰ 30° 590 N 410 N 30° G -17.5 m E 30 220 N

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Q: The truss pinned to the floor at D, and supported by a roller at point A is loaded as shown in the…

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Q: Determine the ratio of the factored compressive load and the design compressive strength.0…

A: Answer  Calculation of ratio of factored compressive load and the design compressive strength For…

Q: Q4/ The rigid beam is supported by a pin at A and two steel wires BD and CE. Each rod has a diameter…

A: Given Esteel=200 Gpa  a= 1 m, b= 1.5 m, d= 40 mm, h= 10 mm

Q: for sew A the a bove and lim pinjointed and joints. Stiffness also la portal frame, tixed at at 'D',…

A: Frame : A frame is a structure which has combination of beam, column and slab which resist the…

Q: Determine the resultant moment of forces about the point O? Neglect the beam thickness *KN 15m #2kN…

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Q: The frame is subjected to a load of 5 kN which acts on member ABD at D. Determine the required…

A: Given : Load = 5 kN Failure shear stress = 50 MPa FOS = 1.5 Determine the required dia. of pin D…

Q: A simply supported wood beam that is 13.2 m long carries a 41 kN concentrated load at B. The…

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Q: The steel truss was subjected to horizontal force 2kN at C. Find the m of AC using the method of…

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Q: The roof of a single-story storage building shown is subjected to a uniformly distributed load of…

A: Given Data: UDL=2.4kPah=6mL=8m Required- To find the maximum bending moment (kN-m) in beam BE of a…

Q: Q4/ The rigid beam is supported by a pin at A and two steel wires BD and CE. Each rod has a diameter…

A: E=200GPa

Q: Problem Number 4. The steel truss was subjected to horizontal force 2kN at C. Find the member forces…

A: P = 2KN

Q: A tubular post of outer diameter d₂ is guyed by two cables fitted with turnbuckles (see figure).…

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Q: A simply supported rectangular concrete beam 6m long is reinforced with four-28mm diam  tension bars…

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Q: #1 Consider the compound beam shown. Suppose that P₁= 1000 N, P₂ = 1600 N, w = 600 N/m and point E…

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Q: Determine the maximum shear stress acting in the beam at the section where the internal shear force…

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Q: The beam of triangular cross-section has a maximum allowable compressive stress of 175 MPa. Take the…

A: Answer We can calculate the Maximum Allowable Bending Moment using Bending equation. From Bending…

Q: H.W Replace the loading equivalent resultant force and specify its location on the beam, measured…

A: Consider the given beam. The resultant force will be equal to the algebraic sum of all the forces.

Q: The steel truss was subjected to horizontal force 2kN at C. Find the member forces of AC using the…

A: Calculation of Member forces in AC using Unit Load METHOD##

Q: F9-5. The beam is subjected to the load at its end Determine the maximum principal stress at point…

A: Given : Determine the maximum principle stress at point B ? Concept -  Principle stresses are taken…

Q: Determine the force in member CG for the truss using the method of joints. 133 kN 67 kN B F G -4 at…

A: Given Data  Find the force in member CG (FCG) using method of joint.

Q: If the beam shown is supported by the fixed wall at B and the rod AC. if the rod has a cross…

A: Using compatibility equations

Q: 1, determine the magnitude of the shear stress in the beam at point 1, determine the magnitude of…

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Q: Determine the external reactions at A and F for the roof truss loaded as shown (positive answers are…

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Q: 370 N y 240 N B 60° Answers: 410 N 60 30° 530 N 540 N 30° G -10.5 m 30° 250 N

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Q: Determine the resultant force at pins A, B, and C on the three-member frame. P=129 N q=200 N/m

A: Load P = 129 NLoad intensity q = 200 N/m

Q: 1. Compute the force in member FEF 2. Compute the force in member FED

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Q: Situation 1: From the truss shown, each member has a cross-sectional area of A = 400 mm 2, E = 200…

A: Area of each member, A = 400mm2Young's modulus, E = 200GPaThe truss is loaded with a downwards point…

Q: A simply supported rectangular concrete beam 6m long is reinforced with four-28mm diam tension bars…

A: The values given in the question are as follows: Length of beam(L)=6 m Tension reinforcement=4-28 mm…

Q: Thumbs-Up will be given. Write the complette solutions legibly. Answer in 3 decimal places…

A: Solution: Given: DL = 6 kNm LL = 9 kNm Factored Load: P=1.2 DL + 1.6 LL=1.2×6+1.6×9P=21.6 kNm DL = 6…

Q: A 3 470 B

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Q: The number of significant digits is set to 3. The tolerance is +-1 in the 3rd significant digit. The…

A: The given data is shown below: Mass of beam=410 kg

Q: a negligible mass m block. It is pinne of 20 kg and negli static friction at B-

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Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.