This is AI generated Response.
Asked by: Md Israrul Haque
Created at: 2025-10-18 10:28:10
Final Answer:
RenderedRaw Text
Did you like this answer?
# Solution
The following solution is presented in a digital format, preserving all the information from the handwritten pages.
---
## Given Data and Diagram
A structure is shown with points A, C, and D:
- Vertical and horizontal distances are marked:
- From A to C: horizontal 1 m, vertical 3 m
- From C to D: horizontal 3 m, vertical .5 m
- From D vertically down: 2 m
- Loads: 4 kN at A, 6 kN at D
---
## At Point A: Moment Equilibrium
At point A,
\[
\sum M_A =
\]
### The Free Body Diagram at \(\sum M_A = \):
- Forces at D: \(T_{CD}\)
- Decompose \(T_{CD}\) using geometry:
- \(T_{CD} \left(\frac{4}{\sqrt{17}}\right)4 + T_{CD} \left(\frac{1}{\sqrt{17}}\right)2 - 6(4) - 4(1) = \)
Using diagram values:
\[
3.8806T_{CD} + .4851T_{CD} - 28 =
\]
### Solve for \(T_{CD}\):
\[
T_{CD} = \frac{28}{3.8806 + .4851} = 6.414\, \text{kN}
\]
So,
\[
T_{CD} = 6.414\, \text{kN}
\]
---
## At Joint C: Force Equilibrium
Next, analyzing joint C:
\[
\sum F_x = \quad \text{(right direction)}
\]
Let’s resolve forces at joint C:
- \(T_{CD}\) acts along CD
- Unknown: \(T_{BC}\)
---
## Solving for \(T_{BC}\):
- \(T_{BC} = 1.571\)
- \(\theta = 8.14^\circ\)
---
## For Joint AB:
\[
T_{AB}\sin\phi = 1.571\sin(8.14^\circ) + 4
\]
\[
T_{AB}\sin\phi = 3.7776 \quad \text{(Equation 4)}
\]
\[
T_{AB}\cos\phi = 1.5552 \quad \text{(Equation 3)}
\]
\[
T_{AB}\sin\phi = 3.7776 \quad \text{(Equation 5)}
\]
Solving equations 3 and 5:
\[
\phi = \tan^{-1}\left(\frac{3.7776}{1.5552}\right) = 67.62^\circ
\]
\[
T_{AB} = \frac{3.7776}{\sin\phi} = \frac{3.7776}{\sin(67.62^\circ)} = 4.085\, \text{kN}
\]
---
## Maximum Tensile Force
The maximum tensile force is:
\[
T_{CD} = 6.414\, \text{kN}
\]
(Compare above three cables.)
---
## Geometry Calculation
From geometry:
\[
\tan\phi = \frac{y_B}{1}
\]
\[
y_B = 1 \cdot \tan(67.62^\circ)
\]
\[
y_B = 2.429\, \text{m}
\]
---
## Summary Diagram
- In the triangle formed by A, B, and D:
- \(y_B = 2.429\, \text{m}\)
- \(\phi = 67.62^\circ\)
---
All steps and calculations have been carefully transcribed from the provided solution, preserving the mathematical process and reasoning.
VIPSolutions focuses on clear, step‑by‑step explanations so you can learn quickly. Whether you need an AI‑generated walkthrough or a short hint to get unstuck, each solution is organized for fast reading and easy review later.
Search similar questions, compare approaches, and bookmark the best answers for revision. Our goal is simple: quick, reliable study help that feels natural—not noisy.
VIPSolutions
i Solubon-
ST 1 f
\ i
Cw
\
| RL)
| a
| fen
i yd
AE point £MA -0.
i The Plow dragram at ZL Naso
] AY i
" Pred Ted (iz) Tel TE) -
4 EEO ETHER
. Heve Foilow the dfagram Value,
im NT 3.8806 Tep +0 4st Top “28 =,
ne
Here find out he Teo Value. i
Ch
2 “Hm
Teo = 28 = buy kn
2.8204 + 0,48t] Lat ups co A]
or
Neat we bind oub Jotak ¢
£Fx =o" 5" (Rrght) divecHon.
T TD »
3 Ae vere Find sub the Tp = 9
_ el oo
t b
TAB sno = s+ SEs y re
Tagstag 2m —@ (Tzesisu
- - bass 6-2)
Tap (of ¢ = $582, PX
TAQ {inp = 21717 — ©
By oluing @ %® )
2.7177
” -t ff — > .
P= tan ( rl). REx
Tag = 217% 2m
¥ ———— TUDES kN.
Sno $0 (6742)
The maxtmum ensrle free 38 Teo = bunk
(ompore above Hhyee cable]
Prom He geometry | a
Tang =
vo Ir
Ye z | tan [/)
C
roe enc
Ce