a) Transform the beam into one made entirely of steel. b) The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and determine the maximum stress in the wood if the beam is subjected to a moment of M=1.2kN.m Est = 200GPa, Ew 12GPa. (please give explanations and step by step solutions if possible)

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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a) Transform the beam into one made entirely of steel.

b) The steel channel is used to reinforce the wood beam. Determine the maximum stress in the
steel and determine the maximum stress in the wood if the beam is subjected to a moment of M=1.2kN.m Est = 200GPa, Ew
12GPa.

(please give explanations and step by step solutions if possible)

The steel channel is used to reinforce the wood beam. Determine the maximum stress in the
steel and the wood if the beam is subjected to a moment of M=1.2kN.m Est = 200GPa, Ew =
12GPa.
12mm
A
3.'S. IO.
****
SI
TEMININING
FEMININ...
***
12mm
INI
MINI
SEUR
5.1.3.
TE
2277
***
.......
----
175mm
100mm
12mm
1.2kN.m
Transcribed Image Text:The steel channel is used to reinforce the wood beam. Determine the maximum stress in the steel and the wood if the beam is subjected to a moment of M=1.2kN.m Est = 200GPa, Ew = 12GPa. 12mm A 3.'S. IO. **** SI TEMININING FEMININ... *** 12mm INI MINI SEUR 5.1.3. TE 2277 *** ....... ---- 175mm 100mm 12mm 1.2kN.m
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Can you please provide greater detail on how we got the value of 199 and a diagram for the solution.

Let us evaluate the y
Σ TA
ΣΑ
у
=
I
y =
y = 86.11 mm
Now
Moment of inertia I
199x12³
12
12×199×6+2×((.
×( ( 1¹75 – 12 )×12×88 ) +10.5×( ¹75 – 1/2)×88
2
2
-
12
12×199+2× ( ¹75 — 12 )×12+10.5×( ¹75 – ¹2² )
2
2
24x883
+ 24 × 88
12
+ 10.5 × 12 × (94 – 86. 11)
+ 199 x 12 x (86. 11 − 6) +
× (( ¹75 – 1¹/2) – 86. 11) + 10.5X12³
12
I = 15.89 x 105 mm4
Transcribed Image Text:Let us evaluate the y Σ TA ΣΑ у = I y = y = 86.11 mm Now Moment of inertia I 199x12³ 12 12×199×6+2×((. ×( ( 1¹75 – 12 )×12×88 ) +10.5×( ¹75 – 1/2)×88 2 2 - 12 12×199+2× ( ¹75 — 12 )×12+10.5×( ¹75 – ¹2² ) 2 2 24x883 + 24 × 88 12 + 10.5 × 12 × (94 – 86. 11) + 199 x 12 x (86. 11 − 6) + × (( ¹75 – 1¹/2) – 86. 11) + 10.5X12³ 12 I = 15.89 x 105 mm4
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