NewSchool of Architecture and Design Structural Systems Problems

Anonymous

Question Description

please show work

please make it clear to read and organized

Problem #1

DRAW THE FREE BODY DIAGRAM,
THE SHEAR DIAGRAM, AND THE MOMENT DIAGRAM FOR THE FOLLOWING BEAMS. IDENTIFY THE MAXIMUM SHEAR AND THE MAXIMUM MOMENT FOR EACH CASE

Problem #2

CALCULATE THE MAXIMUM BEAM DEFLECTIONS FOR THE FOLLOWING SITUATIONS, IDENTIFY WHERE ON THE BEAM THE MAXIMUM DEFLECTION OCCURS. (for problem C just identify the deflection at the middle of the span)

Tutor Answer

brianreyes958
School: UCLA

Hello please find the attached PDF and Ms WORD version of the answer.

Problem #1
DRAW THE FREE BODY DIAGRAM,
THE SHEAR DIAGRAM, AND THE MOMENT DIAGRAM FOR THE FOLLOWING
BEAMS. IDENTIFY THE MAXIMUM SHEAR AND THE MAXIMUM MOMENT FOR
EACH CASE

Free body diagram

Summation of all the forces acting horizontally
โˆ‘ ๐น๐‘ฅ = 0

๐‘กโ„Ž๐‘ข๐‘ 

๐ด๐‘ฅ = 0

Summation of all the forces acting vertically
โˆ‘ ๐น๐‘ฆ = 0

๐‘กโ„Ž๐‘ข๐‘ 

๐ด๐‘ฆ + ๐ต๐‘ฆ โˆ’ 5 โˆ’ 5 = 0
๐ด๐‘ฆ + ๐ต๐‘ฆ = 10

(1)

Taking moments about A
โˆ‘ ๐‘š๐ด = 5 โˆ— 5 + 5 โˆ— 10 โˆ’ ๐ต๐‘ฆ โˆ— 15 = 0
๐ต๐‘ฆ =

25 + 50
= 5๐‘˜๐‘–๐‘๐‘ 
15

๐ด๐‘ฆ + 5 = 10 ๐‘กโ„Ž๐‘ข๐‘  ๐ด๐‘ฆ = 5๐‘˜๐‘–๐‘๐‘ 

Shear force
๐‘†๐น๐ด = 5๐‘˜๐‘–๐‘๐‘  ๐‘Ž๐‘›๐‘‘ ๐‘†๐น๐ถ = 5 โˆ’ 5 = 0๐‘˜๐‘–๐‘๐‘ 
๐‘†๐น๐ท = 5 โˆ’ 5 โˆ’ 5 = โˆ’5๐‘˜๐‘–๐‘๐‘ 
๐‘†๐น๐ต = 5 โˆ’ 5 โˆ’ 5 + 5 = 0๐‘˜๐‘–๐‘๐‘ 
Shear force diagram
5โ€™

5โ€™

5โ€™

5 kips
๐ด๐‘ฆ

C

๐ต๐‘ฆ
๐ท
-5 kips

๐‘€๐‘Ž๐‘ฅ๐‘–๐‘š๐‘ข๐‘š ๐‘ โ„Ž๐‘’๐‘Ž๐‘Ÿ ๐‘ ๐‘ก๐‘Ÿ๐‘’๐‘ ๐‘  = 5๐‘˜๐‘–๐‘๐‘ 
Bending moments at point A
๐ต๐‘š๐ด = ๐ด๐‘ฆ โˆ— 0 = 0 ๐‘˜๐‘–๐‘๐‘ . ๐‘“๐‘ก
๐ต๐‘š๐ถ = ๐น๐ถ โˆ— 5 = 25 ๐‘˜๐‘–๐‘๐‘ . ๐‘“๐‘ก
๐ต๐‘š๐ท = ๐น๐ท โˆ— 10 โˆ’ ๐น๐ถ โˆ— 5 = 50 โˆ’ 25 = 25 ๐‘˜๐‘–๐‘๐‘ . ๐‘“๐‘ก
๐ต๐‘š๐ต = ๐ต๐‘ฆ โˆ— 15โˆ’๐น๐ท โˆ— 10 โˆ’ ๐น๐ถ โˆ— 5 = 50 โˆ’ 25 = 0 ๐‘˜๐‘–๐‘๐‘ . ๐‘“๐‘ก
๐‘€๐‘Ž๐‘ฅ๐‘–๐‘š๐‘ข๐‘š ๐‘๐‘’๐‘›๐‘‘๐‘–๐‘›๐‘” ๐‘š๐‘œ๐‘š๐‘’๐‘›๐‘ก = 25 ๐‘˜๐‘–๐‘๐‘ . ๐‘“๐‘ก
Bending moment diagram
5โ€™

5โ€™

5โ€™

25 kips.ft
๐‘š๐‘ฅ

C

๐ท

B)

Summation of all the forces acting horizontally
โˆ‘ ๐น๐‘ฅ = 0

๐‘กโ„Ž๐‘ข๐‘ 

๐ด๐‘ฅ = 0

Summation of all the forces acting vertically
โˆ‘ ๐น๐‘ฆ = 0

๐‘กโ„Ž๐‘ข๐‘  ๐ด๐‘ฆ โˆ’ 10 โˆ’ 5 โˆ’ 5 = 0 ๐‘กโ„Ž๐‘ข๐‘  ๐ด๐‘ฆ = 15๐‘˜๐‘–๐‘๐‘ 

Taking moments about A
โˆ‘ ๐‘š๐ด = โˆ’10 โˆ— 5 โˆ’ 5 โˆ— 20 = 200๐‘˜๐‘–๐‘๐‘ . ๐‘“๐‘ก
Shear force
๐‘†๐น๐ด = ๐ด๐‘ฆ = 15๐‘˜๐‘–๐‘๐‘  ๐‘Ž๐‘...

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