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For the given cross section of the fit the shape board, with dimensions W' by L',...
Need the last 4 parts please The moment of inertia about the x axis from the...
Need the last 4 parts please
The moment of inertia about the x axis from the entire cross-section. 20/20 pts 669 ft4 System Answer: 669 The moment of inertia around the y axis due to section B. 20/20 pts 61.667 ft4 ✓ System Answer: 61.666666666667 The moment of inertia about the y axis from the entire cross-section. 20 / 20 pts 1891 System Answer: 1891 The product moment of inertia due to section C. ENTER 3 tries remaining. 10 point(s)...
w lb/ft bft Cross section CC' l ft at B. W. rPoint Course A The horizontal...
w lb/ft bft Cross section CC' l ft at B. W. rPoint Course A The horizontal reaction at A. b ENTER 3 tries remaining. 1 pointis) possible The vertical reaction at A lb ENTER 3 tries remaining. 1 points) possible The range for section 1. t ENTER 3 tries remaining. 1 point's) possible The shear force equation for section 1. b ENTER 3 tries remaining. 1 point/s) possible The bending moment equation for section 1. b.ft ENTER 3 tries remaining....
A composite shape is constructed by two angles and a plate to a W shape as...
A composite shape is constructed by two angles and a plate to a W shape as shown. Determine the y-centroid the moment of inertia with respect to the x-axis ar thebase y and the y-axis (). Using the parallel-axis theorem -Ls (Similar to Problem 9.51 distance o-bar) for the composite shape shown. I+A ), compute the moment of inertia parallel with the x-axis at the centroid (.). Al units are in inches. The shape is symmetrical about the 5 in.-...
w lb/ft Wlb M Ib-ft hft b ft Cross section EE l, ft ft 2 W....
w lb/ft Wlb M Ib-ft hft b ft Cross section EE l, ft ft 2 W. a m FourPoint Course The vertical reaction at B. b ENTER 3 tries remaining. 1 point's) possibie The range for section 1. ft ENTER 3 tries remaining. 1 pointis) possible The shear force equation for section 1 b ENTER 3 tries remaining. 1 pointis) possible The bending moment equation for section 1. b.ft ENTER 3 tries remaining. 1 poinns) possible The range for section...
Parallel-Axis Theorem for an Area 2 of 8 Learning Goal: I, Iy = ft To be...
Parallel-Axis Theorem for an Area 2 of 8 Learning Goal: I, Iy = ft To be able to use the parallel-axis theorem to calculate the moment of inertia for an area. The parallel-axis theorem can be used to find an area's Submit axis that passes through the centroid and whose moment of inertia is known. If ar and y' are the axes that pass through an area's centroid, the parallel-axis theorem for the moment about the x axis, moment about...
Consider the beam below: w lb/ft Wlb h ft M lb-ft ---H --- bft Cross section...
Consider the beam below: w lb/ft Wlb h ft M lb-ft ---H --- bft Cross section EE li ft * COLLAPSE IMAGES W = 53 lb/ft = 148 lb ly = 8 ft 2 = 5 ft b = 0.5 ft h = 0.1 ft M= 143 lb.ft The vertical reaction at A. ENTER 3 tries remaining. 1 point(s) possible The horizontal reaction at A. lb ENTER 3 tries remaining. 1 point(s) possible The vertical reaction at B. ENTER 3...
For the beam show, calculate: w lb/ft p lb/ft L ft aft bft COLLAPSE IMAGES w...
For the beam show, calculate: w lb/ft p lb/ft L ft aft bft COLLAPSE IMAGES w = 80 lb/ft p = 40 lb/ft L = 8 ft a = 8 ft b= 6 ft The vertical reaction at A. Іь ENTER 3 tries remaining. 1 point(3) possible The vertical reaction at B. lb ENTER 3 tries remaining. 1 point(3) possible The range for section 1. ) ft ENTER 3 tries remaining. 1 point(3) possible The shear force equation for section...
Q3 Weight: 20.00% Consider the beam below: w lb/ft Wlb h ft M lb-ft w ---HI...
Q3 Weight: 20.00% Consider the beam below: w lb/ft Wlb h ft M lb-ft w ---HI bft Cross section EE COLLAPSE IMAGES W = 62 lb/ft W= 139 lb /y = 7 ft 12 =2 ft b = 0.5ft h = 0.1 ft M = 127 lb.ft int.com/#!/assessment/2015/247/Homework%2013 The vertical reaction at A. lb ENTER 3 tries remaining. 1 points) possible The horizontal reaction at A 1/1 pts 0 lb System Answer 0 The vertical reaction at B. lb ENTER...
For the beam shown, draw the reactions at supports A and B in the positive direction,...
For the beam shown, draw the reactions at supports A and B in
the positive direction, and also draw the shear and bending moment
in the positive direction on your FBD.
where w=23 kip/ft L=5 ft
find
w kip ft C A B L/3 ft L ft The shear equation across the beam. kip ENTER X 2 tries remaining. 1 point(s) possible The bending moment equation across the beam. kip.ft ENTER 3 tries remaining. 1 point(s) possible The internal shear...
For the beam shown, draw the reactions at supports A and B in the positive direction,...
For the beam shown, draw the reactions at supports A and B in the positive direction, and also draw the shear and bending moment in the positive direction on your FBD. w lb/ft aft L ft aft COLLAPSE IMAGES w = 480 lb/ft L = 13 ft 9 = 2 ft Calculate the support reaction Ay lb ENTER 3 tries remaining. 1 point(s) possible Calculate the support reaction Ax lb ENTER 3 tries remaining. 1 point(3) possible Calculate the support...
Need the last 4 parts please
The moment of inertia about the x axis from the entire cross-section. 20/20 pts 669 ft4 System Answer: 669 The moment of inertia around the y axis due to section B. 20/20 pts 61.667 ft4 ✓ System Answer: 61.666666666667 The moment of inertia about the y axis from the entire cross-section. 20 / 20 pts 1891 System Answer: 1891 The product moment of inertia due to section C. ENTER 3 tries remaining. 10 point(s)...
w lb/ft bft Cross section CC' l ft at B. W. rPoint Course A The horizontal reaction at A. b ENTER 3 tries remaining. 1 pointis) possible The vertical reaction at A lb ENTER 3 tries remaining. 1 points) possible The range for section 1. t ENTER 3 tries remaining. 1 point's) possible The shear force equation for section 1. b ENTER 3 tries remaining. 1 point/s) possible The bending moment equation for section 1. b.ft ENTER 3 tries remaining....
A composite shape is constructed by two angles and a plate to a W shape as shown. Determine the y-centroid the moment of inertia with respect to the x-axis ar thebase y and the y-axis (). Using the parallel-axis theorem -Ls (Similar to Problem 9.51 distance o-bar) for the composite shape shown. I+A ), compute the moment of inertia parallel with the x-axis at the centroid (.). Al units are in inches. The shape is symmetrical about the 5 in.-...
w lb/ft Wlb M Ib-ft hft b ft Cross section EE l, ft ft 2 W. a m FourPoint Course The vertical reaction at B. b ENTER 3 tries remaining. 1 point's) possibie The range for section 1. ft ENTER 3 tries remaining. 1 pointis) possible The shear force equation for section 1 b ENTER 3 tries remaining. 1 pointis) possible The bending moment equation for section 1. b.ft ENTER 3 tries remaining. 1 poinns) possible The range for section...
Parallel-Axis Theorem for an Area 2 of 8 Learning Goal: I, Iy = ft To be able to use the parallel-axis theorem to calculate the moment of inertia for an area. The parallel-axis theorem can be used to find an area's Submit axis that passes through the centroid and whose moment of inertia is known. If ar and y' are the axes that pass through an area's centroid, the parallel-axis theorem for the moment about the x axis, moment about...
Consider the beam below: w lb/ft Wlb h ft M lb-ft ---H --- bft Cross section EE li ft * COLLAPSE IMAGES W = 53 lb/ft = 148 lb ly = 8 ft 2 = 5 ft b = 0.5 ft h = 0.1 ft M= 143 lb.ft The vertical reaction at A. ENTER 3 tries remaining. 1 point(s) possible The horizontal reaction at A. lb ENTER 3 tries remaining. 1 point(s) possible The vertical reaction at B. ENTER 3...
For the beam show, calculate: w lb/ft p lb/ft L ft aft bft COLLAPSE IMAGES w = 80 lb/ft p = 40 lb/ft L = 8 ft a = 8 ft b= 6 ft The vertical reaction at A. Іь ENTER 3 tries remaining. 1 point(3) possible The vertical reaction at B. lb ENTER 3 tries remaining. 1 point(3) possible The range for section 1. ) ft ENTER 3 tries remaining. 1 point(3) possible The shear force equation for section...
Q3 Weight: 20.00% Consider the beam below: w lb/ft Wlb h ft M lb-ft w ---HI bft Cross section EE COLLAPSE IMAGES W = 62 lb/ft W= 139 lb /y = 7 ft 12 =2 ft b = 0.5ft h = 0.1 ft M = 127 lb.ft int.com/#!/assessment/2015/247/Homework%2013 The vertical reaction at A. lb ENTER 3 tries remaining. 1 points) possible The horizontal reaction at A 1/1 pts 0 lb System Answer 0 The vertical reaction at B. lb ENTER...
For the beam shown, draw the reactions at supports A and B in
the positive direction, and also draw the shear and bending moment
in the positive direction on your FBD.
where w=23 kip/ft L=5 ft
find
w kip ft C A B L/3 ft L ft The shear equation across the beam. kip ENTER X 2 tries remaining. 1 point(s) possible The bending moment equation across the beam. kip.ft ENTER 3 tries remaining. 1 point(s) possible The internal shear...
For the beam shown, draw the reactions at supports A and B in the positive direction, and also draw the shear and bending moment in the positive direction on your FBD. w lb/ft aft L ft aft COLLAPSE IMAGES w = 480 lb/ft L = 13 ft 9 = 2 ft Calculate the support reaction Ay lb ENTER 3 tries remaining. 1 point(s) possible Calculate the support reaction Ax lb ENTER 3 tries remaining. 1 point(3) possible Calculate the support...
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