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Problem No.4: For the following beam, draw the Influence lines for: a) reaction at A, b)...
Using STATICS DRAW the INFLUENCE LINES for the VERTICAL REACTION AT "A"
(Influence Line Creation) Problem 1. Using STATICS DRAW the INFLUENCE LINES for the VERTICAL REACTION AT "A", the SHEAR AT "E", and the MOMENT AT "E". SHOW ALL VALUES. Assume the support 'A' is a pin, "B" and "D" are rollers, and "C" is a hinge. (12 points) FREE BODY DIAGRAMS & CALCULATIONS
(Influence Line Use) Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE...
(Influence Line Use) Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE PLACEMENT OF THE LOADS AND CALCULATE THE VALUES FOR THE MAXIMUM NEGATIVE SHEAR at "B", and MAXIMUM POSITIVE MOMENT at "B" due to a concentrated point load of 1400 lb, uniform live load of 800 lb/ft, and a uniform dead load of 600 lb/ft. (12 points) B с 10 ft 6 ft 2A 2 ft Vs ww 2.25 IL RA -0.25 1.25 0.25 IL...
1) Draw the influence lines for the reaction force at A, the shear force at C...
1) Draw the influence lines for the reaction force at A, the shear force at C and the moment at C. Show all the values. faA 2m 2 m 2 m 2) The Beam of problem #1 is subjected to uniform dead load 0.8 kN/m and uniform live load of 2 kN/m and a single live concentrated force of 20 kN. Determine (a) the maximum reaction force at A, (b) the maximum positive shear at point C, and (c) the...
1. (40 pts.) HingeHinge 6 m 6 m 6 m 6 m Consider the given continuous beam above and a) Use Müller Breslau Principle and draw the influence lines of the vertical support reaction at C, shear force...
1. (40 pts.) HingeHinge 6 m 6 m 6 m 6 m Consider the given continuous beam above and a) Use Müller Breslau Principle and draw the influence lines of the vertical support reaction at C, shear force at B and moment at B. Calculate the ordinates at the points A, B, C, D, E, F and G. (30 pts.) e the maximum positive shear force at B considering the following loading below. Also show what will be the loading:...
(Influence Line Use) Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE...
(Influence Line Use) Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE PLACEMENT OF THE LOADS AND CALCULATE THE VALUES FOR THE MAXIMUM NEGATIVE SHEAR at "B", and MAXIMUM POSITIVE MOMENT at "B" due to a concentrated point load of 1400 lb, uniform live load of 800 lb/ft, and a uniform dead load of 600 lb/ft. (12 points) B с 10 ft 6 A 2A 2 ft Vs. 2.25 IL RA -0.25 1.25 0.25 IL VB...
PROBLEMS 6-1. Draw the influence lines for (a) the moment at C, (b) the reaction at...
PROBLEMS 6-1. Draw the influence lines for (a) the moment at C, (b) the reaction at B, and (c) the shear at C. Assume A is pinned and B is a roller. Solve this problem using the basic method of Sec. 6–1. 3 m 3 m - - - 3 m Probs. 6-1/6-2
Question 2: A simply supported beam under loading as shown in Figure 1: 1. Draw the influence lines of the bending moment and shear force at point C (L/4) Using the influence lines to determine t...
Question 2: A simply supported beam under loading as shown in Figure 1: 1. Draw the influence lines of the bending moment and shear force at point C (L/4) Using the influence lines to determine the bending moment and shear force at section C due to the loading as shown in the figure. 2. 3. There is a distributed live load (w#2.5kN/m) which can vary the location along the beam. Determine the location of the live loads which create the...
AP2: Beam AD is connected to a cable at C.: a) Draw the influence lines for the force in cable CE b) Draw the influence...
AP2: Beam AD is connected to a cable at C.: a) Draw the influence lines for the force in cable CE b) Draw the influence lines for the vertical reaction at support A c) Draw the influence lines for the moment at EB d) Determine the maximum absolute value of each response quantity from parts a) to c) if beam AD supports a 15-ton concentrated load and a movable/variable 1 kip/ft uniformly distributed live load. 6 4 43'
Problem #2 6-7. Draw the influence lines for (a) the moment at B (b) the shear...
Problem #2 6-7. Draw the influence lines for (a) the moment at B (b) the shear at B, and (c) the vertical reaction at A. Solve this problem using the basic method of Sec. 6.1. Hint: The support at C resists only a horizontal force and a bending moment.
Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE PLACEMENT OF THE...
Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE PLACEMENT OF THE LOADS AND CALCULATE THE VALUES FOR THE MAXIMUM NEGATIVE SHEAR at "B", and MAXIMUM POSITIVE MOMENT at "B" due to a concentrated point load of 1400 lb, uniform live load of 800 lb/ft, and a uniform dead load of 600 lb/ft. (12 points) B с А 10 ft 6 24 2 ft Vs. 2.25 IL RA -0.25 1.25 0.25 IL VB -0.25 -0.75 1.5...
(Influence Line Use) Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE PLACEMENT OF THE LOADS AND CALCULATE THE VALUES FOR THE MAXIMUM NEGATIVE SHEAR at "B", and MAXIMUM POSITIVE MOMENT at "B" due to a concentrated point load of 1400 lb, uniform live load of 800 lb/ft, and a uniform dead load of 600 lb/ft. (12 points) B с 10 ft 6 ft 2A 2 ft Vs ww 2.25 IL RA -0.25 1.25 0.25 IL...
1) Draw the influence lines for the reaction force at A, the shear force at C and the moment at C. Show all the values. faA 2m 2 m 2 m 2) The Beam of problem #1 is subjected to uniform dead load 0.8 kN/m and uniform live load of 2 kN/m and a single live concentrated force of 20 kN. Determine (a) the maximum reaction force at A, (b) the maximum positive shear at point C, and (c) the...
1. (40 pts.) HingeHinge 6 m 6 m 6 m 6 m Consider the given continuous beam above and a) Use Müller Breslau Principle and draw the influence lines of the vertical support reaction at C, shear force at B and moment at B. Calculate the ordinates at the points A, B, C, D, E, F and G. (30 pts.) e the maximum positive shear force at B considering the following loading below. Also show what will be the loading:...
(Influence Line Use) Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE PLACEMENT OF THE LOADS AND CALCULATE THE VALUES FOR THE MAXIMUM NEGATIVE SHEAR at "B", and MAXIMUM POSITIVE MOMENT at "B" due to a concentrated point load of 1400 lb, uniform live load of 800 lb/ft, and a uniform dead load of 600 lb/ft. (12 points) B с 10 ft 6 A 2A 2 ft Vs. 2.25 IL RA -0.25 1.25 0.25 IL VB...
PROBLEMS 6-1. Draw the influence lines for (a) the moment at C, (b) the reaction at B, and (c) the shear at C. Assume A is pinned and B is a roller. Solve this problem using the basic method of Sec. 6–1. 3 m 3 m - - - 3 m Probs. 6-1/6-2
Question 2: A simply supported beam under loading as shown in Figure 1: 1. Draw the influence lines of the bending moment and shear force at point C (L/4) Using the influence lines to determine the bending moment and shear force at section C due to the loading as shown in the figure. 2. 3. There is a distributed live load (w#2.5kN/m) which can vary the location along the beam. Determine the location of the live loads which create the...
AP2: Beam AD is connected to a cable at C.: a) Draw the influence lines for the force in cable CE b) Draw the influence lines for the vertical reaction at support A c) Draw the influence lines for the moment at EB d) Determine the maximum absolute value of each response quantity from parts a) to c) if beam AD supports a 15-ton concentrated load and a movable/variable 1 kip/ft uniformly distributed live load. 6 4 43'
Problem #2 6-7. Draw the influence lines for (a) the moment at B (b) the shear at B, and (c) the vertical reaction at A. Solve this problem using the basic method of Sec. 6.1. Hint: The support at C resists only a horizontal force and a bending moment.
Problem 2. The INFLUENCE LINES ARE GIVEN for the beam below. DRAW THE PLACEMENT OF THE LOADS AND CALCULATE THE VALUES FOR THE MAXIMUM NEGATIVE SHEAR at "B", and MAXIMUM POSITIVE MOMENT at "B" due to a concentrated point load of 1400 lb, uniform live load of 800 lb/ft, and a uniform dead load of 600 lb/ft. (12 points) B с А 10 ft 6 24 2 ft Vs. 2.25 IL RA -0.25 1.25 0.25 IL VB -0.25 -0.75 1.5...
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