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Member end shears and moments for drawing V and M diagrams (designer's sign convection):

Member end shears and moments for drawing V and M diagrams (designer's sign convection):

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Please help me with finding the V max M max negative and M max positive and the sketch the deflected shape of the beam.

sign convention): 4 k-ft rk Help App -1ok.0 -12k-, 3.5k Google Play -4K-H mework ee Answers -12-1t Member end shears and moments for drawing V and M diagrams (designer's sign convection): t Gear is in made to 45k Iklft Hey, stuck with your writing assignments? Park St 4 kft -I2 k-1t -12t- birections lok -3. 4.5 (k) (k) (k) F12 M. (k-) 4 (k-1) 4 delete (k-t) } X- 2-12 12 t2 m= 10,1 k XIA DE
Transcribed Image Text:sign convention): 4 k-ft rk Help App -1ok.0 -12k-, 3.5k Google Play -4K-H mework ee Answers -12-1t Member end shears and moments for drawing V and M diagrams (designer's sign convection): t Gear is in made to 45k Iklft Hey, stuck with your writing assignments? Park St 4 kft -I2 k-1t -12t- birections lok -3. 4.5 (k) (k) (k) F12 M. (k-) 4 (k-1) 4 delete (k-t) } X- 2-12 12 t2 m= 10,1 k XIA DE
Sketch the deflected shape of the beam. nternal hape k/ft 図 12 ft fom TITTT 4' 4' 6' 4. E=30,000 ksi (= 4,320,000 ksf) I=400 in* (=0.0193 ft*)
Transcribed Image Text:Sketch the deflected shape of the beam. nternal hape k/ft 図 12 ft fom TITTT 4' 4' 6' 4. E=30,000 ksi (= 4,320,000 ksf) I=400 in* (=0.0193 ft*)
Expert Solution
Step 1

There is an internal hinge at 4. So, we can split the given beam into two parts. One is left part of the beam and another is right part of the beam. 

First of all, consider the right part of the beam. The free body diagram of the right part of the beam is

Mechanical Engineering homework question answer, step 1, image 1

where, R4 and R5 are the reactions at 4 and 5

Take a moment about 5

R4×6=12R4=2 k

Equate the vertical forces

R5 = R4 = 2 k

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