agrams for the beam shown below. 10 kN p= 20 kN/m

Q: Compute the maximum bending moment of the given beam below. (kN-m)

A: Using trial and error to find zero shear positionFIND REACTIOIN AT B:∑MD=0…

Q: using three-moment equation

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Q: 1. Draw the shear and bending moment diagrams for the beams shown. Determine the maximum values of…

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Q: 1. (a) Slope at A; (b) deflection at midpsan. 6 kN/m 1 kN-m 5m B

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Q: a) Compute for the moment of Inertia of the cross section b) Determine the maximum bending stress in…

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Q: or the frame below, if the axial force in column AE is 9.18 k, what is the approximate axial force…

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Q: for the simply supported beam shown. The = (maximum value of shear force (V 2m 50 KN 10 m 50 kN 2m…

A: Given : simply supported beam Load W= 50 kN at  2m from support To determine: maximum shear force

Q: A simple beam AB is loaded by two segments of uniform load and two horizontal and vertical forces…

A: The free body diagram of the beam is shown below:

Q: Draw a single FBD that would allow you to solve for the force in members DG, DF, and EF knowing that…

A: Determine the force in members DG, DF and EF. Let's solve using method of section.

Q: The beam shown is loaded with the triangular, rectangular and the 10kN point load. Find the…

A: Given- A simply supported beam carrying triangular load, rectangular load and point load. Required-…

Q: in 79 A W250 x 49.1 section (with the web vertical) is used as a cantilever beam 4-m-long. The beam…

A: Given Data:Section W250×49.1 UDL=woIt is asked To determine: σmaxζmax.

Q: Calculate the maximum Vu and Mu in the beam shown below: 2m P(D)=20 KN P(L)=10 KN 4m 3. B

A: Given,  Simply supported beam with point load    To find : Maximum Shear force (Vu) and maximum…

Q: 3.5 (A). A beam AB, 3 m long, is simply-supported at A and E. It carries a 16 kN concentrated load…

A: From Equilibrium Condition ∑ Fy = 0 ,  RA + RB = 25KN   ∑ MA = 0  , 16×1.2 + 1.8×5×2.1 = RB×3 RB =…

Q: 1. For the restrained beam shown below, compute the end moments (kN-m). 1 m 3 kN-m 1 m 1. What is M…

A: we will use slope deflection method to determine the Moment at the ends

Q: Please answer fast

A: The answers have been attached in the form of images: If you have any doubt feel free to ask in the…

Q: 15 kN/m 10 kN/m A C 2.5EI 1.5EI 3 m 3 m

A: Solution:- Deflection in the beam AB, δAB=5×wL4384EI        =5×10×34384×2.5EI      =4.218EI…

Q: 4) Determine the resultant force and specify where it acts on the beam measured from A. Provide a…

A: Consider the free body diagram of the given beam as shown below. Consider the summation of external…

Q: A continuous beam is loaded as shown 18. Which of the following gives the fixed end moment AB in…

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Q: OMENT(M) AND SKETCH THE SHEAR FORCE AND BENDING MOMENT DIAGRAMS. NEGLECT THE WEIGHT OF THE BEAM. 3.)…

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Q: Determine the shear and moment equations of the beam. Illustrate the shear and moment diagrams.…

A: Let's divide total load into two part for easy calculation.  Part one having uniformly distributed…

Q: A wooden beam 150 mm. wide by 300 mm. deep is loaded as shown. Allowable bending stress is 8 MPa. 1.…

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Q: Find the deflection at the center and slope at the ends for a beam 8m long simply supported at ends…

A: E = 200 GN/m2 = 2×108 kN/m2 I = 10-4 m4 Length of span, L = 8m UDL, w = 4 kN/m Concentrated load, P…

Q: hid-span

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Q: For the beam shown, determine the following: a. Maximum Positive Shear b. Maximum Negative Moment c.…

A: Given data:

Q: Draw S.F and B.M diagrams for the beam loaded as shown below. Use graphical method. -1.5 kN/m 0.5…

A: Solution  Given diagram

Q: 25. Which of the follo value of Ay? а. 25 KN b. -40 KN 26. Which of the follo value of Cx? 60 kN -30…

A: To find the support reactions (Ay , Bz and Cx) -

Q: A beam with a solid homogeneous rectangular section is simply supported at A and B. A concentrated…

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Q: In the propped beam shown, Solve for the following 1. wall moment 2. maximum deflection .....

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Q: 5x102 kN-m2. 2 KN 2 W- 5 kN/m

A: Given Data: Overgaiging beam Length L=12mUDL all over the span = 5kN/mPoint Loat at Left and Right…

Q: The shear diagram for a beam is given below. 30 kN 25 kN -10 kN -20 kN 25 m+is m-t--2 m m -2.5 m…

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Q: If the magnitude of the 50kN point load is replaced by 65 kN and the magnitude of 70 kN/m…

A: Given data,  Point load = 65 kN UDL = 83 kN/m UVL = 90 kN/m at maximum and 0 at starting.

Q: Two forces and a couple are applied to the beam in an attempt to lift it. Replace this force system…

A: Given data:

Q: For the beam loaded as shown in the figure below, determine the following unknov method. а. Moment…

A: Solution:MA1=-wl212=24.3×8212=-129.6 kN.mMC1=wl212=24.3×8212=129.6 kN.m As per the given Figure:

Q: What will be the depth of a beam that is 20 cm wide and is subjected to shearing force of 10,000 kg.…

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System of Forces

A force is an external influence or agent that changes the state of a system. According to Newton's second law, the force applied to a body is proportional to the change in momentum of a body. Hence, whenever a force is applied to a body, its momentum changes. If the body is at rest, under the action of force, its state of rest changes to motion. In other words, the magnitude of the momentum of the body changes from zero to a measurable value. Similarly, if the body possesses some initial velocity, it has some associated momentum, the effect of force can cause the body to halt or come to rest, that is, the momentum reduces to zero.