 # NewSchool of Architecture and Design Structural Systems Problems Anonymous

### Question Description

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)

### Unformatted Attachment Preview

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
∑ 𝐹𝑥 = 0

𝑡ℎ𝑢𝑠

𝐴𝑥 = 0

Summation of all the forces acting vertically
∑ 𝐹𝑦 = 0

𝑡ℎ𝑢𝑠

𝐴𝑦 + 𝐵𝑦 − 5 − 5 = 0
𝐴𝑦 + 𝐵𝑦 = 10

(1)

∑ 𝑚𝐴 = 5 ∗ 5 + 5 ∗ 10 − 𝐵𝑦 ∗ 15 = 0
𝐵𝑦 =

25 + 50
= 5𝑘𝑖𝑝𝑠
15

𝐴𝑦 + 5 = 10 𝑡ℎ𝑢𝑠 𝐴𝑦 = 5𝑘𝑖𝑝𝑠

Shear force
𝑆𝐹𝐴 = 5𝑘𝑖𝑝𝑠 𝑎𝑛𝑑 𝑆𝐹𝐶 = 5 − 5 = 0𝑘𝑖𝑝𝑠
𝑆𝐹𝐷 = 5 − 5 − 5 = −5𝑘𝑖𝑝𝑠
𝑆𝐹𝐵 = 5 − 5 − 5 + 5 = 0𝑘𝑖𝑝𝑠
Shear force diagram
5’

5’

5’

5 kips
𝐴𝑦

C

𝐵𝑦
𝐷
-5 kips

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 = 5𝑘𝑖𝑝𝑠
Bending moments at point A
𝐵𝑚𝐴 = 𝐴𝑦 ∗ 0 = 0 𝑘𝑖𝑝𝑠. 𝑓𝑡
𝐵𝑚𝐶 = 𝐹𝐶 ∗ 5 = 25 𝑘𝑖𝑝𝑠. 𝑓𝑡
𝐵𝑚𝐷 = 𝐹𝐷 ∗ 10 − 𝐹𝐶 ∗ 5 = 50 − 25 = 25 𝑘𝑖𝑝𝑠. 𝑓𝑡
𝐵𝑚𝐵 = 𝐵𝑦 ∗ 15−𝐹𝐷 ∗ 10 − 𝐹𝐶 ∗ 5 = 50 − 25 = 0 𝑘𝑖𝑝𝑠. 𝑓𝑡
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 25 𝑘𝑖𝑝𝑠. 𝑓𝑡
Bending moment diagram
5’

5’

5’

25 kips.ft
𝑚𝑥

C

𝐷

B)

Summation of all the forces acting horizontally
∑ 𝐹𝑥 = 0

𝑡ℎ𝑢𝑠

𝐴𝑥 = 0

Summation of all the forces acting vertically
∑ 𝐹𝑦 = 0

𝑡ℎ𝑢𝑠 𝐴𝑦 − 10 − 5 − 5 = 0 𝑡ℎ𝑢𝑠 𝐴𝑦 = 15𝑘𝑖𝑝𝑠

∑ 𝑚𝐴 = −10 ∗ 5 − 5 ∗ 20 = 200𝑘𝑖𝑝𝑠. 𝑓𝑡
Shear force
𝑆𝐹𝐴 = 𝐴𝑦 = 15𝑘𝑖𝑝𝑠 𝑎�...

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