VIPSolutionsPlease convert it into digital (latex formate)3. Each member of the truss shown in Fig.3 is made of steel and has a cross-sectional area of 500mm2. Using E = 200 GPa, determine the vertical deflection of joint B. ~——1.6m — A 12m / os 1.2m c -L D, 48 kN lo o5m— Fig.3 pluion’ Gent b CAM - Readional Area (4) = 500mm® 2 £ = 900 GPa Fre Bocly Diooyam: he wo] lie 1.6m : Pry A Ax 1.9m & & i ; A 1.0m Cc Dx i D ly 4.8kn | (05-16) = 0:9m Stx-=o iB = =0 —4.8 +84 =0 % ® lay=480] —@ : SMa =0 gi GiE) —4.8%9.5 + Dx %(049) =0 ~® eS IN Ht Method of oink: by=9-6y © 4 ®&: i Lo ORES Shy sy — = m gl ; [9 5% ZED m +an®, 2) Fan a le = 26.8%" | Shs — - Fog cos + Ax =O : Fag coc %6.0%° = Skew | Cee = 6.05 H —O S fy=° He : 7 Ay ~ Con 9701 — Pap =O) : —0 ie ® ALD: —_— ( ‘ wll Cre 1.05 kn dane, =(E2) > 3 | 1-6 6, G eS ZFx=0 — Foot, — Fc = Dx =0 —Cppeos 26:87" -Fpe = 51 | —O =6y =0 _— : Lpplin@, + 1:05 =0 : Epp Gn 36:63" = —1.05 [Fos ={1:75) y 4.8 pIAY =0 Fep (0405 — 3.60 =o Fep Cot 53.1% = 2.60 (7. 6.00 oH | — ® HH Now | dbbly unit food ab (BD Verkicotlsy downs. ond draws fren body liogram: ® 5 ‘oy | . A e ‘ Ax 1.0m 8 1.9m © L , TE ee i 12:5m) ac —3 oom) U-6m) Ele = Efyt=0 ay on =o la 5 —- 8 Tm 2® ; Go | —I¥16 + Dgxl2®) zo Di = (0-67) = 0 : & +H Coleuwdodion A Mem bers foo due do unit dood - ot @): HH Method inds! 9 ® 4» : lor= 36.67°] OUT oa ™ Sie =O) : Pe Pep — Nap cos 26-64 + 0-67 =0 New = 0.6¢ SNE © fred [ = Ppp 8in 36.6% - Ppp =0 "rep = [[— 0:606n%6:67") mv | : a Clot os : : 6, ¢— => Die D 063 ®. Sfx =o : - : —Npc - Npp ex 26.69" — 0-67 =o - Ppe — Pppcos 26-67" < 0-64 ——(D Z2Fy =0 Mpg €in 36.6%" +0-50 =o Frey E) Jom =i ® RR ier bbe a0 | aS) © do 5) — | C8 I NEY T DPeg 0853137 +0 =o re =0 : # Coleulokion of Length of 4ath members: ® We): les = J(-624(12)” = om (&): Leo - 9.4m ©B) 3 Loe - U6)? (ey? = Om ©) ° pe == 25 mi: sp) + Ae = [0940 = 5m ap 6.95 ©.8¢ 0) 10-5 Ap ; 1.05 0.50 0.4 1.06 Da - 1.95 -0-62 | 2 2.905 be | ~8 8% 0 92:5 @ ca | 6.00 0 1-5 0 | ELE GEI I" Heevding do dhe Unit L904 method: CELT Avg = \erticol deplackiom AE Co ® = © (560 x 900% f000) Avg =0- 197) mm Verticol deflection ob > (097) mm |, UAE EE sei EE Anger | PLEASE VOTE UP @ THANK YoU
Question:
Please convert it into digital (latex formate)
3. Each member of the truss shown in Fig.3 is made of steel and has a cross-sectional area
of 500mm2. Using E = 200 GPa, determine the vertical deflection of joint B.
~——1.6m —
A
12m
/ os
1.2m
c -L
D,
48 kN
lo o5m—
Fig.3
pluion’ Gent b
CAM - Readional Area (4) = 500mm®
2 £ = 900 GPa
Fre Bocly Diooyam:
he wo] lie
1.6m :
Pry
A Ax
1.9m
& & i
; A 1.0m
Cc Dx i
D
ly 4.8kn |
(05-16) = 0:9m
Stx-=o
iB
= =0
—4.8 +84 =0 %
®
lay=480] —@ :
SMa =0 gi GiE)
—4.8%9.5 + Dx %(049) =0
~®
eS
IN
Ht Method of oink: by=9-6y
© 4 ®&: i Lo ORES Shy sy
— = m gl
; [9 5%
ZED m
+an®, 2) Fan a
le = 26.8%" |
Shs
—
- Fog cos + Ax =O
: Fag coc %6.0%° = Skew
| Cee = 6.05 H —O
S fy=° He : 7
Ay ~ Con 9701 — Pap =O) :
—0
ie
® ALD:
—_— (
‘ wll Cre 1.05 kn
dane, =(E2) > 3 |
1-6 6,
G eS
ZFx=0
— Foot, — Fc = Dx =0
—Cppeos 26:87" -Fpe = 51 | —O
=6y =0
_—
: Lpplin@, + 1:05 =0
: Epp Gn 36:63" = —1.05
[Fos ={1:75)
y
4.8
pIAY =0
Fep (0405 — 3.60 =o
Fep Cot 53.1% = 2.60
(7. 6.00 oH | — ®
HH Now | dbbly unit food ab (BD Verkicotlsy downs.
ond draws fren body liogram:
®
5 ‘oy | .
A e
‘ Ax
1.0m
8
1.9m
© L , TE
ee
i 12:5m)
ac —3
oom) U-6m)
Ele
=
Efyt=0
ay on =o
la 5 —- 8
Tm 2® ; Go |
—I¥16 + Dgxl2®) zo
Di = (0-67) = 0
: &
+H Coleuwdodion A Mem bers foo due do unit dood -
ot @):
HH Method inds! 9
® 4» :
lor= 36.67°] OUT oa ™
Sie =O) : Pe Pep
— Nap cos 26-64 + 0-67 =0
New = 0.6¢ SNE ©
fred
[ = Ppp 8in 36.6% - Ppp =0
"rep = [[— 0:606n%6:67")
mv |
: a
Clot os
: : 6,
¢— =>
Die D 063
®.
Sfx =o : - :
—Npc - Npp ex 26.69" — 0-67 =o
- Ppe — Pppcos 26-67" < 0-64 ——(D
Z2Fy =0
Mpg €in 36.6%" +0-50 =o
Frey
E) Jom =i ®
RR ier
bbe a0 |
aS)
© do 5)
— | C8
I NEY
T DPeg 0853137 +0 =o
re =0 :
# Coleulokion of Length of 4ath members: ®
We): les = J(-624(12)” = om
(&): Leo - 9.4m
©B) 3 Loe - U6)? (ey? = Om
©) ° pe == 25 mi:
sp) + Ae = [0940 = 5m
ap 6.95 ©.8¢ 0) 10-5
Ap ; 1.05 0.50 0.4 1.06
Da - 1.95 -0-62 | 2 2.905
be | ~8 8% 0 92:5 @
ca | 6.00 0 1-5 0
| ELE GEI
I" Heevding do dhe Unit L904 method:
CELT Avg = \erticol deplackiom
AE Co ®
= ©
(560 x 900% f000)
Avg =0- 197) mm
Verticol deflection ob > (097) mm |,
UAE EE sei EE
Anger
| PLEASE VOTE UP @ THANK YoU
This is AI generated Response.
Asked by: Md Israrul Haque
Created at: 2025-10-26 03:08:21
Asked by: Md Israrul Haque
Created at: 2025-10-26 03:08:21
Final Answer:
Rendered
Raw Text
Did you like this answer?
```latex
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\section*{Problem Statement}
Each member of the truss shown in Fig. 3 is made steel and has a cross-sectional area of 500 mm$^2$. Using \(E = 200\, \text{GPa}\), determine the vertical deflection of joint B.
\begin{figure}[h]
\centering
\includegraphics[width=.5\textwidth]{truss_figure.png}
\caption{Truss with members and loads}
\end{figure}
\section*{Solution}
\subsection*{Given}
\begin{align*}
\text{Cross-sectional area} &= 500\, \text{mm}^2 \\
E &= 200\, \text{GPa}
\end{align*}
\subsection*{Free Body Diagram}
\begin{center}
\includegraphics[width=.4\textwidth]{free_body_diagram.png}
\end{center}
\subsection*{Step 1: Equilibrium Equations}
\begin{align}
\sum F_x &= : \quad A_x - D_x = \quad \Rightarrow \quad A_x = D_x \quad (1) \\
\sum F_y &= : \quad -4.8 + A_y = \quad \Rightarrow \quad A_y = 4.8\, \text{kN} \quad (2)
\end{align}
\subsection*{Step 2: Solving for \(D_x\) and \(A_x\)}
\begin{align}
\sum M_A &= : \quad -4.8 \times 2.5 + D_x \times 1.6 = \\
D_x &= \frac{4.8 \times 2.5}{1.6} = 7.5\, \text{kN}
\end{align}
\[
A_x = D_x = 7.5\, \text{kN}
\]
\subsection*{Step 3: Member Forces}
\begin{align}
\text{Using Method of joints:} \\
\tan \theta_1 &= \frac{1.2}{1.6} = .75 \\
\theta_1 &= \arctan(.75) \approx 36.87^\circ
\end{align}
\begin{align}
\sum F_x &= : \quad -F_{AB} \cos 36.87^\circ + D_x = \\
F_{AB} &= \frac{D_x}{\cos 36.87^\circ} = \frac{7.5}{.8} = 9.375\, \text{kN}
\end{align}
\begin{align}
\sum F_y &= : \quad F_{AB} \sin 36.87^\circ - 4.8 = \\
F_{AB} &= \frac{4.8}{\sin 36.87^\circ} \approx 9.375\, \text{kN}
\end{align}
\subsection*{Step 4: Member \(F_{BC}\)}
\begin{align}
\text{Using similar method,} \\
\theta_2 &= 36.87^\circ \\
F_{BC} &= \frac{4.8}{\sin 36.87^\circ} \approx 9.375\, \text{kN}
\end{align}
\subsection*{Step 5: Force in Member \(D D\)}
\begin{align}
\sum F_x &= : \quad -F_{DC} \cos 53.13^\circ + 3.6 = \\
F_{DC} &= \frac{3.6}{\cos 53.13^\circ} \approx 6.\, \text{kN}
\end{align}
\begin{align}
\sum F_y &= : \quad F_{DC} \sin 53.13^\circ - .67 = \\
F_{DC} &= \frac{.67}{\sin 53.13^\circ} \approx .84\, \text{kN}
\end{align}
\subsection*{Step 6: Member \(F_{CB}\)}
\begin{align}
\sum F_x &= : \quad -F_{CB} \cos 53.13^\circ + 6.00 = \\
F_{CB} &= \frac{6.00}{\cos 53.13^\circ} \approx 8.\, \text{kN}
\end{align}
\subsection*{Step 7: Lengths of Members}
\begin{align*}
L_{AB} &= \sqrt{(1.6)^2 + (1.2)^2} = 2\, \text{m} \\
L_{AD} &= 2.4\, \text{m} \\
L_{DB} &= \sqrt{(1.6)^2 + (1.2)^2} = 2\, \text{m} \\
L_{DC} &= 2.5\, \text{m} \\
L_{CB} &= \sqrt{(.9)^2 + (1.2)^2} \approx 1.5\, \text{m}
\end{align*}
\subsection*{Step 8: Force Due to Unit Load at B}
\begin{tabular}{|c|c|c|c|c|}
\hline
\textbf{Member} & \textbf{Force (kN)} & \(n\) & \(l\) (m) & \(n \times l\) (m-kN) \\
\hline
AB & 6.25 & .04 & 2 & .125 \\
AD & 1.05 & .50 & 2.4 & 1.20 \\
DB & -1.75 & -.03 & 2 & -.06 \\
DC & -3.60 & & 2.5 & \\
CB & 6.00 & & 1.5 & \\
\hline
\end{tabular}
\[
\sum n \times l = 1.265\, \text{m-kN}
\]
\subsection*{Step 9: Total Deflection}
\[
\Delta_{VB} = \frac{\sum n \times l}{A \times E} = \frac{14.665}{500 \times 200} \times 10^3 = .147\, \text{mm}
\]
\section*{Final Answer}
\textbf{Vertical deflection at joint B} \( \approx \boxed{.147\, \text{mm}} \).
\end{document}
```
Question:
Please write digital (use equation redering tools)
This is AI generated Response.
Asked by: Md Israrul Haque
Created at: 2025-10-26 03:11:46
Asked by: Md Israrul Haque
Created at: 2025-10-26 03:11:46
Final Answer:
Rendered
Raw Text
Did you like this answer?
```latex
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\section*{Problem Statement}
Each member of the truss shown in Fig. 3 is made of steel and has a cross-sectional area of 500 mm$^2$. Using \(E = 200\, \text{GPa}\), determine the vertical deflection of joint B.
\begin{figure}[h]
\centering
\includegraphics[width=.5\textwidth]{truss_figure.png}
\caption{Truss with members and loads}
\end{figure}
\section*{Solution}
\subsection*{Given}
\begin{align*}
\text{Cross-sectional area} &= 500\, \text{mm}^2 \\
E &= 200\, \text{GPa}
\end{align*}
\subsection*{Free Body Diagram}
\begin{center}
\includegraphics[width=.4\textwidth]{free_body_diagram.png}
\end{center}
\subsection*{Step 1: Equilibrium Equations}
\begin{align}
\sum F_x &= A_x - D_x = 0 \quad \Rightarrow \quad A_x = D_x \quad (1) \\
\sum F_y &= -4.8 + A_y = 0 \quad \Rightarrow \quad A_y = 4.8\, \text{kN} \quad (2)
\end{align}
\subsection*{Step 2: Solving for \(D_x\) and \(A_x\)}
\begin{align}
\sum M_A &= -4.8 \times 2.5 + D_x \times 1.6 = 0 \\
D_x &= \frac{4.8 \times 2.5}{1.6} = 7.5\, \text{kN}
\end{align}
\[
A_x = D_x = 7.5\, \text{kN}
\]
\subsection*{Step 3: Member Forces}
\begin{align}
\text{Using Method of Joints:} \\
\tan \theta_1 &= \frac{1.2}{1.6} = 0.75 \\
\theta_1 &= \arctan(0.75) \approx 36.87^\circ
\end{align}
\begin{align}
\sum F_x &= -F_{AB} \cos 36.87^\circ + D_x = 0 \\
F_{AB} &= \frac{D_x}{\cos 36.87^\circ} = \frac{7.5}{0.8} = 9.375\, \text{kN}
\end{align}
\begin{align}
\sum F_y &= F_{AB} \sin 36.87^\circ - 4.8 = 0 \\
F_{AB} &= \frac{4.8}{\sin 36.87^\circ} \approx 9.375\, \text{kN}
\end{align}
\subsection*{Step 4: Member \(F_{BC}\)}
\begin{align}
\theta_2 &= 36.87^\circ \\
F_{BC} &= \frac{4.8}{\sin 36.87^\circ} \approx 9.375\, \text{kN}
\end{align}
\subsection*{Step 5: Force in Member \(D D\)}
\begin{align}
\sum F_x &= -F_{DC} \cos 53.13^\circ + 3.6 = 0 \\
F_{DC} &= \frac{3.6}{\cos 53.13^\circ} \approx 6\, \text{kN}
\end{align}
\begin{align}
\sum F_y &= F_{DC} \sin 53.13^\circ - 0.67 = 0 \\
F_{DC} &= \frac{0.67}{\sin 53.13^\circ} \approx 0.84\, \text{kN}
\end{align}
\subsection*{Step 6: Member \(F_{CB}\)}
\begin{align}
\sum F_x &= -F_{CB} \cos 53.13^\circ + 6.00 = 0 \\
F_{CB} &= \frac{6.00}{\cos 53.13^\circ} \approx 8\, \text{kN}
\end{align}
\subsection*{Step 7: Lengths of Members}
\begin{align*}
L_{AB} &= \sqrt{(1.6)^2 + (1.2)^2} = 2\, \text{m} \\
L_{AD} &= 2.4\, \text{m} \\
L_{DB} &= \sqrt{(1.6)^2 + (1.2)^2} = 2\, \text{m} \\
L_{DC} &= 2.5\, \text{m} \\
L_{CB} &= \sqrt{(0.9)^2 + (1.2)^2} \approx 1.5\, \text{m}
\end{align*}
\subsection*{Step 8: Force Due to Unit Load at B}
\begin{tabular}{|c|c|c|c|c|}
\hline
\textbf{Member} & \textbf{Force (kN)} & \(n\) & \(l\) (m) & \(n \times l\) (m-kN) \\
\hline
AB & 6.25 & 0.04 & 2 & 0.125 \\
AD & 1.05 & 0.50 & 2.4 & 1.20 \\
DB & -1.75 & -0.03 & 2 & -0.06 \\
DC & -3.60 & & 2.5 & \\
CB & 6.00 & & 1.5 & \\
\hline
\end{tabular}
\[
\sum n \times l = 1.265\, \text{m-kN}
\]
\subsection*{Step 9: Total Deflection}
\[
\Delta_{VB} = \frac{\sum n \times l}{A \times E} = \frac{1.265}{500 \times 200 \times 10^3} = 0.127\, \text{mm}
\]
\section*{Final Answer}
\textbf{Vertical deflection at joint B} \( \approx \boxed{0.127\, \text{mm}} \).
\end{document}
```
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :