Please show work and write clearly. Thank you. 6-21. The compound beam is subjected to a...
Up ompound beam is subjected to a uniform dead load of 200 lb/ft and a uniform live load of 150 lb/ft. ermine (a) the maximum negative moment these loads develop at A, and (b) the maximum positive Shear these loads develop at D. Assume B is a hinge. El = constant. S: Live loads may be placed on any segment on the beam to maximize a function. Also, take advantage of the influence lines. There is load no point A...
PartA The compound beam shown in (Figure 1) is subjected to a uniform dead load of 300 lb/ft and a single live load of 2 k. Assume C is a fixed support, B is a pin, and A is a roller Determine the negative bending moment with the maximum magnitude created by these loads at C Express your answer using three significant figures (Mc)max(-) k ft Figure 1 of 1 Submit uest Answer Part B Determine the negative shear with...
The compound beam is subjected to a uniform dead load of 200 lb/ft and a uniform live load of 150 lb/ft. Assume Bis a pin and is a roller. Follow the sign convention (Figure 1) Submit Previous Answers Correct Part B Determine the maximum positive shear at D Express your answer using three significant figures. Figure 1 of 1 ΜΗ ΑΣφ 11 vec ? (V) lb B D Submit Previous Answers Request Answer 51 X Incorrect; Try Again: 2 attempts...
This is an influence line
question, please use that to solve for the max negative moment and
shear. Thank you!
KAssignment 8 (Influence Lines, Values, 2 questions, 6 marks inc. 2 bonus marks) Problem 6.20 Part A The compound beam shown in (Figure 1) is subjected to a uniform dead load of 350 lb/ft and a single live load of 5 k. Assume C is a fixed support, B is a pin, and A is a roller Determine the negative...
Consider the beam subjected to a concentrated load consisting of
2.25 kips of dead load and 5.55 kips of live load at point B. Find
maximum factored beam shear, moment, and deflection.
Consider the beam and loading given below. The beam is subjected to a concentrated load consisting of 2.25 kips of dead load and 5.55 kips of live load at point B. Neglect beam weight. You may use any information from the AISC Manual, a) Draw the general shape...
(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...
(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...
The beam shown in Figure (2) is subjected to a uniform live load of 2.4 kN/m, a dead load of 1.0 kN/m, and a single live load of 80 kN. Assume B is an internal hinge. Determine: (a) the maximum positive moment at E, (b) the maximum positive shear at E. created by these loads. Hinge A B С E D 2m 2m 2m 2m
Q1. The beam supports a uniform dead load of 500 N/m and single
live concentrated force of 3000 N. Determine (a) the maximum
positive moment that can be developed at point C, and (b) the
maximum positive shear that can be developed at point C. Assume the
support at A is a pin and B is a roller.
1. The beam supports a uniform dead load of 500 N/m and single live concentrated force of 3000 N. Determine (a) 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...