Approximate Analysis of
Rectangular Building Frame.
Lecture 18
VERTICAL LOADINGS
ON BUILDING FRAME
APPROXIMATE
ANALYSIS
If the columns are extremely stiff,
no rotation at A and B will occur,
and the deflection for the girder
will look like this
The column supports at A and B will each exert
three reactions on the girder, and therefore the
girder will be statically indeterminate to the
third degree (6 reactions – 3 equations of
equilibrium). To make the girder statically
determinate, an approximate analysis will
therefore require three assumptions
APPROXIMATE ANALYSIS
Assumption
• There is zero moment in
the girder, 0.1L from the
left support.
• There is zero moment in
the girder, 0.1L from the
right support.
• The girder does not
support an axial force.
Problem #1
Determine approximately the support action at A, B and C
Problem #1
Determine approximately the support
action at A, B and C
𝐹𝑜𝑟 𝐹𝐺
6 𝑘𝑁/𝑚
4.5𝑚
6 𝑘𝑁/𝑚
𝐹𝑦 = 𝐺𝑦 =
3.6 𝑚
𝐹𝑦
𝑀𝐹
6 𝑘𝑁/𝑚
6𝑥3.6
= 10.8 𝑘𝑁
2
𝑅𝐹 = 𝑅𝐺 = 10.8 + 6𝑥0.45 = 13.5 𝑘𝑁
𝐺𝑦
6 𝑘𝑁/𝑚
𝑀𝐺
𝑀𝐹 = 𝑀𝐺 = 10.8𝑥0.45 + 6𝑥0.45 0.45/2
0.45 𝑚
𝑅𝐹
0.45 𝑚
𝑅𝐺
𝑀𝐹 = 𝑀𝐺 = 5.47 𝑘𝑁𝑚
Determine approximately the
support action at A, B and C
Problem #1
𝐹𝑜𝑟 𝐺𝐻
6 𝑘𝑁/𝑚
6𝑚
6 𝑘𝑁/𝑚
6𝑥4.8
𝐺𝑦 = 𝐻𝑦 =
= 14.4 𝑘𝑁
2
4.8 𝑚
𝐺𝑦
𝑀𝐺
𝐻𝑦
6 𝑘𝑁/𝑚
6 𝑘𝑁/𝑚
0.6 𝑚
0.6 𝑚
𝑅𝐺
𝑅𝐻
𝑀𝐻
𝑅𝐺 = 𝑅𝐻 = 14.4 + 6𝑥0.6 = 18 𝑘𝑁
𝑀𝐺 = 𝑀𝐻 = 14.4𝑥0.6 + 6𝑥0.6 0.6/2
𝑀𝐺 = 𝑀𝐻 = 9.72 𝑘𝑁𝑚
Problem #1
Determine approximately the
support action at A, B and C
𝐹𝑜𝑟 𝐷𝐸
18 𝑘𝑁/𝑚
𝐷𝑦 = 𝐸𝑦 =
4.5 𝑚
18𝑥3.6
= 32.4 𝑘𝑁
2
18 𝑘𝑁/𝑚
𝑅𝐷 = 𝑅𝐸 = 32.4 + 18𝑥0.45 = 40.5 𝑘𝑁
3.6 𝑚
𝐷𝑦
𝑀𝐸
18 𝑘𝑁/𝑚
𝐸𝑦
18 𝑘𝑁/𝑚
𝑀𝐸
𝑀𝐷 = 𝑀𝐸 = 32.5𝑥0.45 + 18𝑥0.45 0.45/2
𝑀𝐷 = 𝑀𝐸 = 16.44 𝑘𝑁𝑚
0.45 𝑚
𝑅𝐷
0.45 𝑚
𝑅𝐸
Problem #1
Determine approximately the
support action at A, B and C
Problem #1
Determine approximately the
support action at A, B and C
Portal
Frames and truss
Lecture 18
PORTAL FRAMES
• PORTAL FRAMES are frequently used over the entrance
of bridge and as main stiffening element in building
design in order to transfer horizontal forces applied at the
top of the frame to the foundation.
• On bridges, these frames resist the forces caused by the
wind, earthquake, and unbalanced traffic loading on the
bridge deck.
• PORTALS can be pin supported, fixed supported, or
partial fixity.
PIN SUPPORTED
FIXED SUPPORTED
Partial Fixity Frame
Problem#1
Determine (approximately) the force in each truss member of the portal frame. Also find the
reactions at the fixed column supports A and B. Assume all members of the truss to be pin
connected at their ends.
Since the truss frame is fixed supported, we assume the
internal hinge at the center of the column.
Since the truss frame is fixed supported, we assume the
internal hinge at the center of the column.
Problem#1
𝑯
𝑰
Since the truss frame is fixed supported, we assume the
internal hinge at the center of the column.
Problem#1
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑝𝑎𝑟𝑡
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑡ℎ𝑒 ℎ𝑖𝑛𝑔𝑒
2+1
= 1.5 𝑘
2
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑡ℎ𝑒 ℎ𝑖𝑛𝑔𝑒
𝐻𝑥 = 𝐼𝑥 =
+ ෍ 𝑀𝐻 = 0
2𝑥12 + 1𝑥6 − 𝐼𝑦 𝑥16 = 0
𝐼𝑦 = 1.875 𝑘
𝐻𝑦 = 𝐼𝑦 = 1.875 𝑘
𝑴𝐴
𝑴𝑩
Since the truss frame is fixed supported, we assume the
internal hinge at the center of the column.
Problem#1
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝐿𝑜𝑤𝑒𝑟 𝑝𝑎𝑟𝑡
𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑓𝑖𝑥𝑒𝑑 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠
𝐻𝑥 = 𝐴𝑥 = 1.5 𝑘
𝐵𝑥 = 𝐼𝑥 = 1.5 𝑘
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑡ℎ𝑒 ℎ𝑖𝑛𝑔𝑒
𝐻𝑦 = 𝐴𝑦 = 1.875 𝑘
+ ෍ 𝑀𝐴 = 0
1.5 𝑘
𝐼𝑦 = 𝐵𝑦 = 1.875 𝑘
+ ෍ 𝑀𝐵 = 0
1.5 𝑘
𝐻𝑥 𝑥6 − 𝑀𝐴 = 0
𝐼𝑥 𝑥6 − 𝑀𝐵 = 0
𝑀𝐴 = 9 𝑘𝑓𝑡
𝑀𝐵 = 9 𝑘𝑓𝑡
𝑴𝐴
𝑴𝑩
𝐻𝑥 = 𝐴𝑥 = 1.5 𝑘
Since the truss frame is fixed supported, we assume the
internal hinge at the center of the column.
Problem#1
𝐻𝑦 = 𝐴𝑦 = 1.875 𝑘
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑝𝑎𝑟𝑡 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒
𝐺𝐹
𝐺𝐷𝑥
𝐺𝐷𝑦
𝐺𝐷
𝐶𝐷
𝐺𝐷𝑦 = 𝐺𝐷
+ ෍ 𝑀𝐺 = 0
6
1.5 𝑘
𝐻𝑥 𝑥12 − 1𝑥6 − 𝐶𝐷𝑥6 = 0
𝐶𝐷 = 2 𝑘 𝑇
6
10
10
𝐺𝐷𝑥
8
+↑ ෍ 𝐹𝑦 = 0
1.875 𝑘
+ ෍ 𝑀𝐷 = 0
1.5 𝑘
6
−𝐺𝐷
− 𝐻𝑦 = 0
10
1.875 𝑘
𝐺𝐷 = −3.125 𝑘
𝐻𝑥 𝑥6 + 2𝑥6 − 𝐻𝑦 𝑥 8 + 𝐺𝐹𝑥6 = 0
𝐺𝐷 = 3.125 𝑘 𝐶
𝐺𝐹 = 1 𝑘 𝐶
Since the truss frame is fixed supported, we assume the
internal hinge at the center of the column.
Problem#1
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑝𝑎𝑟𝑡 𝑟𝑖𝑔ℎ𝑡 𝑠𝑖𝑑𝑒
𝐻𝑥 = 𝐴𝑥 = 1.5 𝑘
𝐺𝐹 = 1 𝑘
+ ෍ 𝑀𝐹 = 0
𝐹𝐷𝑥
𝐹𝐷
𝐻𝑦 = 𝐴𝑦 = 1.875 𝑘
𝐹𝐷 = 3.125 𝑘
1.5 𝑘
𝐹𝐷𝑦
𝐼𝑥 𝑥12 + 𝐷𝐸𝑥6 = 0
𝐷𝐸
𝐷𝐸 = −3 𝑘
𝐷𝐸 = 3 𝑘 𝐶
6
𝐹𝐷𝑦 = 𝐹𝐷
10
10
𝐹𝐷𝑥
+↑ ෍ 𝐹𝑦 = 0
1.875 𝑘
6
−𝐹𝐷
8
6
+ 𝐼𝑦 = 0
10
𝐹𝐷 = 3.125 𝑘 𝑇
𝐺𝐹 = 1 𝑘 𝐶
𝐺𝐷 = 3.125 𝑘 𝐶
𝐶𝐷 = 2 𝑘 𝑇
𝐴𝑥 = 1.5 𝑘
𝑀𝐴 = 9 𝑘𝑓𝑡
Lateral Loads on
Building FramesPortal Method
The portal method is most suitable for buildings having low elevation and uniform framing.
For the lateral loads acting on portal
frames of fixed support, the points
of inflection occur at approximately
the center of each girder and column
and the columns carry equal shear
loads.
A building bent deflects in the same
way as a portal frame, and therefore
it would be appropriate to assume
that the points of inflection occur at
the center of the columns and
girders as shown.
Considering each bent of the
frame to be composed of a
series of portals, as shown, then
as a further assumption, the
interior
columns
would
represent the effect of two
portal columns and would
therefore carry twice the shear
V as the two exterior columns.
The following are the assumptions for portal method in
analyzing fixed-supported building frames.
1. A hinge is placed at the center of each girder, since
this is assumed to be a point of zero moment.
2. A hinge is placed at the center of each column,
since this is assumed to be a point of zero moment.
3. At a given floor level the shear at the interior
column hinges is twice that at the exterior column
hinges, since the frame is considered to be a
superposition of portals.
Problem#1
Determine (approximately) the reactions at the base of the columns of the frame shown. Use the
portal method of analysis. (Ex. 7.6, RC Hibbeler, 8th ed. p.286)
Problem#1
Determine (approximately) the reactions at the
base of the columns of the frame shown. Use
the portal method of analysis. (Ex. 7.6, RC
Hibbeler, 8th ed. p.286)
෍ 𝐹𝑥 = 0
20 = 𝑉 + 2𝑉 + 𝑉
𝑉 = 5 𝑘𝑁
Problem#1
Determine (approximately) the reactions at
the base of the columns of the frame
shown. Use the portal method of analysis.
(Ex. 7.6, RC Hibbeler, 8th ed. p.286)
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐺
෍ 𝐹𝑥 = 0
20 = 5 + 𝑅𝑥
෍ 𝐹𝑦 = 0
𝑂𝑦 = 𝑅𝑦 = 3.125 𝑘𝑁
𝑅𝑥 = 15 𝑘𝑁
+ ෍ 𝑀𝐺 = 0
5𝑥2.5 − 𝑅𝑦 𝑥4 = 0
𝑅𝑦 = 3.125 𝑘𝑁
Problem#1
Determine (approximately) the reactions at
the base of the columns of the frame shown.
Use the portal method of analysis. (Ex. 7.6,
RC Hibbeler, 8th ed. p.286)
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐺
𝑴𝑮𝑹
+ ෍ 𝑀𝐺 = 0
𝑅𝑦 = 3.125 𝑘𝑁
𝑴𝑮𝑶
𝑀𝐺𝑅 = 𝑅𝑦 𝑥4 = 12.5 kNm
+ ෍ 𝑀𝐺 = 0
𝑀𝐺𝑂 = 5𝑥2.5 = 12.5 kNm
Problem#1
Determine (approximately) the reactions at the base
of the columns of the frame shown. Use the portal
method of analysis. (Ex. 7.6, RC Hibbeler, 8th ed.
p.286)
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐻
෍ 𝐹𝑥 = 0
15 = 10 + 𝑆𝑥
෍ 𝐹𝑦 = 0
𝑃𝑦 = 𝑆𝑦 − 3.125 = 0 𝑘𝑁
𝑆𝑥 = 5 𝑘𝑁
+ ෍ 𝑀𝐻 = 0
10𝑥2.5 − 3.125𝑥4 − 𝑆𝑦 𝑥4 = 0
𝑆𝑦 = 3.125 𝑘𝑁
Problem#1
Determine (approximately) the reactions at the
base of the columns of the frame shown. Use the
portal method of analysis. (Ex. 7.6, RC Hibbeler,
8th ed. p.286)
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐻
𝑴𝑺𝑯
+ ෍ 𝑀𝑆𝐻 = 0
𝑆𝑦 = 3.125 𝑘𝑁
𝑴𝑯𝑹
𝑀𝑆𝐻 = 𝑆𝑦 𝑥4 = 12.5 kNm
𝑴𝑯𝑷
+ ෍ 𝑀𝐻𝑃 = 0
𝑀𝐻𝑃 = 10𝑥2.5 = 25 kNm
+ ෍ 𝑀𝐻𝑅 = 0
𝑀𝐻𝑅 = 3.125𝑥4 = 12.5 kNm
Problem#1
Determine (approximately) the reactions
at the base of the columns of the frame
shown. Use the portal method of analysis.
(Ex. 7.6, RC Hibbeler, 8th ed. p.286)
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐼
+ ෍ 𝑀𝐼𝑆 = 0
𝑀𝐼𝑆 = 3.125𝑥4 = 12.5 kNm
+ ෍ 𝑀𝐼𝑄 = 0
𝑀𝐼𝑄 = 5𝑥2.5 = 12.5 kNm
Problem#1
Determine (approximately) the reactions at the base of the columns of the frame shown. Use the
portal method of analysis. (Ex. 7.6, RC Hibbeler, 8th ed. p.286)
Problem#1
Determine (approximately) the reactions at
the base of the columns of the frame shown.
Use the portal method of analysis. (Ex. 7.6,
RC Hibbeler, 8th ed. p.286)
෍ 𝐹𝑥 = 0
20 + 30 = 𝑉′ + 2𝑉′ + 𝑉′
𝑉′ = 12.5 𝑘𝑁
Problem#1
Determine (approximately) the reactions at
the base of the columns of the frame
shown. Use the portal method of analysis.
(Ex. 7.6, RC Hibbeler, 8th ed. p.286)
+ ෍ 𝑀𝐷𝑀 = 0
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐷
𝑀𝐷𝑀 = 𝑀𝑦 𝑥4 = 50 kNm
+ ෍ 𝑀𝐷𝐽 = 0
𝑀𝐷𝐽 = 12.5𝑥3 = 37.5 kNm
+ ෍ 𝑀𝐷𝑂 = 0
𝑀𝐷𝑂 = 5𝑥2.5 = 12.5 kNm
Problem#1
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐸
+ ෍ 𝑀𝐸𝑁 = 0
𝑀𝐸𝑁 = 12.5𝑥4 = 50 kNm
+ ෍ 𝑀𝐸𝐾 = 0
𝑀𝐸𝐾 = 25𝑥3 = 75 kNm
+ ෍ 𝑀𝐸𝑃 = 0
𝑀𝐸𝑃 = 10𝑥2.5 = 25 kNm
+ ෍ 𝑀𝐸𝑀 = 0
𝑀𝐸𝑀 = 12.5𝑥4 = 50 kNm
Problem#1
Determine (approximately) the reactions at
the base of the columns of the frame
shown. Use the portal method of analysis.
(Ex. 7.6, RC Hibbeler, 8th ed. p.286)
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 𝐹
+ ෍ 𝑀𝑁𝐹 = 0
𝑀𝑁𝐹 = 12.5𝑥4 = 50 kNm
+ ෍ 𝑀𝐹𝐿 = 0
𝑀𝐹𝐿 = 12.5𝑥3 = 37.5 kNm
+ ෍ 𝑀𝐹𝑄 = 0
𝑀𝐹𝑄 = 5𝑥2.5 = 12.5 kNm
Problem#1
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 A
Determine (approximately) the reactions at
the base of the columns of the frame shown.
Use the portal method of analysis. (Ex. 7.6,
RC Hibbeler, 8th ed. p.286)
𝐹𝑜𝑟 𝐽𝑜𝑖𝑛𝑡 B
Problem#1
Determine (approximately) the reactions at
the base of the columns of the frame shown.
Use the portal method of analysis. (Ex. 7.6,
RC Hibbeler, 8th ed. p.286)
𝑴𝑮𝑹
𝑴𝑮𝑶
𝑴𝑯𝑹
𝑴𝑯𝑷
𝑴𝑺𝑯
Lateral Loads on
Building FramesCantilever Method
CANTILEVER METHOD
ASSUMPTIONS
• An inflection points is located
at the middle of each member
of the frame.
• On each storey of the frame, the
axial stress in columns are
linearly proportional to their
distances from the centroid of
the cross sectional areas of all
the columns on that storey.
Problem#2
Use Cantilever method to determine the shear, axial and
moment at A.
Use Cantilever method to determine the shear, axial
and moment at A.
Problem#2
σ 𝑥𝐴
ු
0𝑥10 + 20𝑥8 + 35𝑥6 + 60𝑥10
𝑥ҧ =
=
σ𝐴
10 + 8 + 6 + 10
= 28.53 𝑓𝑡
Use Cantilever method to determine the shear, axial
and moment at A.
Problem#2
28.53 𝑓𝑡
First we will consider the section through hinges at L,M, N and O.
Use Cantilever method to determine the shear, axial
and moment
at A.
First we will
consider the section through hinges at L,M, N and O.
Problem#2
relate the column stresses by proportional
triangles. Expressing the relations in terms of the
force in each column,
𝐹
𝜎=
𝐴
𝐹
𝜎=
𝐴
𝜎𝐿
𝜎𝑀
=
28.53 8.53
𝜎𝐿
𝜎𝑁
=
28.53 6.47
𝐿𝑦
𝑀𝑦
𝐴1
𝐴
= 2
28.53 8.53
𝑁𝑦
𝐿𝑦
𝐴
𝐴1
= 3
28.53 6.47
𝐿𝑦
𝑀𝑦
10 = 8
28.53 8.53
𝐿𝑦
𝑁𝑦
10 = 6
28.53 6.47
𝑀𝑦 = 0.239𝐿𝑦
𝑁𝑦 = 0.136𝐿𝑦
𝜎𝐿
𝜎𝑂
=
28.53 31.47
𝐿𝑦
𝑂𝑦
𝐴1
𝐴4
=
28.53 31.47
𝐿𝑦
𝑂𝑦
10 = 10
28.53 31.47
𝑂𝑦 = 1.103 𝐿𝑦
+ ෍ 𝑀𝑁𝐴 = 0
0.239𝐿𝑦
0.136𝐿𝑦
1.103 𝐿𝑦
8𝑥6 − 𝐿𝑦 𝑥28.53 − 𝑀𝑦 𝑥8.53 − 𝑁𝑦 𝑥6.47 − 𝑂𝑦 𝑥31.47 = 0
𝐿𝑦 = 0.726 𝑘
𝑁𝑦 = 0.99 𝑘
𝑀𝑦 = 0.174 k
𝑂𝑦 = 0.801 k
For column EFGH
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑡𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔
𝐹
𝜎=
𝐴
𝜎𝐸
𝜎𝐹
=
28.53 8.53
𝜎𝐸
𝜎𝐺
=
28.53 6.47
𝜎𝐸
𝜎𝐻
=
28.53 31.47
𝐸𝑦
𝐹𝑦
𝐴1
𝐴
= 2
28.53 8.53
𝐺𝑦
𝐸𝑦
𝐴
𝐴1
= 3
28.53 6.47
𝐸𝑦
𝐻𝑦
𝐴1
𝐴4
=
28.53 31.47
𝐸𝑦
𝐹𝑦
10 = 8
28.53 8.53
𝐸𝑦
𝐺𝑦
10 = 6
28.53 6.47
𝐸𝑦
𝐻𝑦
10 = 10
28.53 31.47
𝐹𝑦 = 0.239𝐸𝑦
𝐺𝑦 = 0.136𝐸𝑦
𝐻𝑦 = 1.103 𝐸𝑦
For column EFGH
𝐴𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑡𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔
+ ෍ 𝑀𝑁𝐴 = 0
0.239𝐸𝑦
0.136𝐸𝑦
1.103 𝐸𝑦
8𝑥20 + (10𝑥8) − 𝐸𝑦 𝑥28.53 − 𝐹𝑦 𝑥8.53 − 𝐺𝑦 𝑥6.47 − 𝐻𝑦 𝑥31.47 = 0
𝐸𝑦 = 3. 627𝑘
𝐹𝑦 = 0.868 k
𝐺𝑦 = 0.494 𝑘
𝐻𝑦 =4. 001 k
For column EFGH
𝟓 𝒌𝑵
𝟏𝟎 𝒌𝑵
Determine (approximately) the force in members GF,GK, and JK of the portal frame. Also find
the reactions at the fixed column supports A and B. Assume all members of the truss to be
connected at their ends.
Use Cantilever method to determine the shear, axial
and moment at A.
Problem#2
Determine the force in each truss
Activity
#14:
Portal
frames
member
of the
portal
frame,
also
compute the reactions at the fixed
column supports A and B.
Assume all members of the truss
to be pin connected at their ends.
Draw the shear and moment
diagram of column DCA.
and truss
Activity
#14:Portal frames and truss
#15: Portal Method-Building
Frame
• Determine the force in each truss
member of the portal frame, also
compute the reactions at the
fixed column supports A and B.
Assume all members of the truss
to be pin connected at their ends.
Draw the shear and moment
diagram of column DCA.
• Determine the approximate
axial forces, shears, and
moments for all the members of
the frames. Draw the moment
diagram