Isolating the right side of section a-a: By summing moments about C, force BD can be determined independently. 1200 lb B 10 BD Resolve the horizontal and vertical components of BD at B. 3. The moment of the vertical component of BD at C is zero. CD 20 V2 +U Mc = 0 3 1200(15) + (BD)(20) = 0 15' C. CE V10 BD = -948.68 lb BD= 948.68 lb (C) 400 lb Since BD turned out to be negative, the assumed sense of this force is incorrect. BD is in compression. Ay = 1200 lb By summing moments at D, force CE can be solved independently. The lines of action of force CD and BD intersect at joint D. 1200 lb B V10 3 +U EMp = -(1200 + 400)(15) + 1200(30) – CE(15) = 0 CE = 800 lb (T) CD 20" 15 C. CE Sum moments at Q to solve for CD. Resolve the horizontal and vertical components of CD at C. The moment of the horizontal Ay=1200 lb 400 lb Solve for distance x by similar triangles: component of CD at Q is zero. 20 1 x = 60 ft +U E MQ -(1200 + 400)(60) + 1200(75) + CD(60) = 0 = 0 CD = 141.42 lb (T) 105

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Chapter2: Loads On Structures
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Can I have some clarification on why the components of the moment of BD became (20)(3)/(10)^1/2 and similarly with CD became (1)(60)/(2)^1/2.? Thank you

Section d-a CUIS The bars BD, CD, and CE. The leti side or ight side Or section d-a can be
isolated as the FBD. Assume tension for the forces BD, CD, and CE.
Isolating the right side of section a-a:
By summing moments about C, force BD can be
determined independently.
1200 lb
10
BD Resolve the horizontal and vertical components of BD at B.
The moment of the vertical component of BD at C is zero.
3
CD
20
V2
+U £Mc = 0
%3D
3
1200(15) +
(BD)(20) = 0
10
%3D
15'
CE
BD = –948.68 lb
BD= 948.68 lb (C)
400 lb
Since BD turned out to be negative, the assumed sense of
this force is incorrect. BD is in compression.
Ay = 1200 lb
By summing moments at D, force CE can be
solved independently. The lines of action of
force CD and BD intersect at joint D.
1200 lb
10 BD
D.
+U £ Mp = 0
-(1200 + 400)(15)+ 1200(30) – CE(15) = 0
CE = 800 lb (T)
CD
%3D
20'
Q
15'
CE
Sum moments at Q to solve for CD. Resolve
the horizontal and vertical components of CD
at C. The moment of the horizontal
X.
Av=1200 lb
400 lb
Solve for distance x by similar
triangles:
component of CD at Q is zero.
3
20
1
+U £Mo = 0
%3D
X- 60 ft
%3D
-(1200 + 400)(60) + 1200(75) +CD(60) = 0
CD = 141.42 lb (T)
%3D
105
C.
Transcribed Image Text:Section d-a CUIS The bars BD, CD, and CE. The leti side or ight side Or section d-a can be isolated as the FBD. Assume tension for the forces BD, CD, and CE. Isolating the right side of section a-a: By summing moments about C, force BD can be determined independently. 1200 lb 10 BD Resolve the horizontal and vertical components of BD at B. The moment of the vertical component of BD at C is zero. 3 CD 20 V2 +U £Mc = 0 %3D 3 1200(15) + (BD)(20) = 0 10 %3D 15' CE BD = –948.68 lb BD= 948.68 lb (C) 400 lb Since BD turned out to be negative, the assumed sense of this force is incorrect. BD is in compression. Ay = 1200 lb By summing moments at D, force CE can be solved independently. The lines of action of force CD and BD intersect at joint D. 1200 lb 10 BD D. +U £ Mp = 0 -(1200 + 400)(15)+ 1200(30) – CE(15) = 0 CE = 800 lb (T) CD %3D 20' Q 15' CE Sum moments at Q to solve for CD. Resolve the horizontal and vertical components of CD at C. The moment of the horizontal X. Av=1200 lb 400 lb Solve for distance x by similar triangles: component of CD at Q is zero. 3 20 1 +U £Mo = 0 %3D X- 60 ft %3D -(1200 + 400)(60) + 1200(75) +CD(60) = 0 CD = 141.42 lb (T) %3D 105 C.
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