How to Analyse Elements in Frames

Introduction

In the last post, we introduced the concept of moment distribution for analyzing structures. However, when analyzing structures it is important for us to adopt a simple but yet methodical approach. By breaking down the structure into a series of segments that can be analyzed discretely the complexity of the analysis is reduced.

In this post, we are going to consider how to break down structures into simpler units that can be analyzed discretely.

Principles

In order to design a structure, the bending moments, shearing forces, axial-forces and torsional moments if present must be determined. The structure can either be analyzed completely as a space frame or broken down into a series of discrete units of subframes. Each of the subframes can then be analyzed and designed individually.

It is normal practice to assume that the supports of continuous beams have no fixity. This is assumption is not strictly true especially for beams framing into columns. It is, therefore, more accurate to analyse them as being part of a frame.

Subframe 1

Subframe one consist of a complete floor beam with the connected columns (fully fixed at the far ends). Analyzing this type of subframe returns the moment and shear in the beam and columns for the floor being considered.

Figure 1 presents sub-frame one, normally a maximum of five span can be analyzed at a time. For larger buildings, a series of overlapping subframes should be used.

Subframe 2

Subframe two: Figure 2 consists of a single span beam and the connecting beam with two adjacent spans all fixed at the far ends. This subframe is used when we are interested in the moment and shear at the central beam. The column moment of the connecting column could also be determined provided that the central beam spans more than the adjacent spans. Figure 2

Subframe 3

Sub-frame 3: Figure 3 can only be used to obtain column moments. We use this type of sub-frame when the analysis of continuous beams have been carried-out using simple supports.

The assumption that the far ends are fixed in subframe 2 & 3 overestimate the stiffness’s of outer beams. For this reason, the stiffness are halved when carrying out analysis as illustrated in Figure 2 & 3.

Elastic Analysis

Any of the traditional methods of analysis method can be used, the moment distribution method is faster and particularly suited for analyzing sub-frames. The frames must be analyzed for all possible load cases in order to determine the critical forces at sections. Load arrangement defined for beams can be used i.e.

All spans carrying the design permanent and variable load, i.e. (1.35Gk + 1.5Qk) and
Alternate spans carrying the design permanent and variable load, i.e. (1.35Gk + 1.5Qk), other spans carrying only the design permanent load, i.e. 1.35Gk

The above-listed load cases will normally be sufficient to yield critical forces at sections. However, if a single load case of maximum load is to be used, the support moments must then be reduced by 20% with consequential increase in the span moments.

The relative stiffness of members may be based on the gross concrete section ignoring reinforcement. For the purpose of calculating the stiffness of flanged beams the flange width of T- and L-beams may be taken from Table 1, in which l = length of the span or cantilever and bw = width of the web.

Table 1: Effective Width of Flanged Beams

Worked Example

Figure 4 is a 3 storey Shopping Mall building having a typical floor plan. The suspended floors are made up of solid-one way slabs and beams. Preliminary sizing of the frame elements have been carried out and is indicated in figure 4. The shopping mall storey heights =3.5m

Beam and columns on gridline C-C will be analyzed using subframe one To illustrate the use of sub-frames in analyzing structures.

ACTIONS

The first step towards analyzing subframe C-C is to determine the actions, to which it will be required to sustain.

1. Permanent Actions:

a)\quad self\quad weight\quad of\quad slab\quad =0.15\times 25=3.75kN/{ m }^{ 2 }

b)\quad finishes\quad and\quad services\quad \quad \quad say\quad \quad =1.5kN/{ m }^{ 2 }

c)\quad partition\quad allowance\quad \quad \quad say\quad \quad \quad \quad =1kN/{ m }^{ 2 }

Permanent\quad Actions\quad ={ g }_{ k }=6.25kN/m^{ 2 }

2. Variable Actions: The floor imposed load has been taken from U.K N.A to BS-EN-1990-1-1

a)\quad floor\quad imposed\quad load\quad \quad \quad =4{ kN/m }^{ 2 }

b)\quad movable\quad partitions\quad say\quad =0.5kN/{ m }^{ 2 }

Variable\quad Actions\quad { q }_{ k }=4.5kN/{ m }^{ 2 }

3. Maximum and Minimum Design Actions on Slab

By observation, the permanent actions is less than 4.5 times the variable actions, therefore equation 6.10b of BS EN 1990-1-1 can be used.

{ n }_{ max }=1.35\xi { g }_{ k }+1.5{ q }_{ k }=(1.35\times 0.925\times 6.25)+(1.5\times 4.5)=14.55kN/{ m }^{ 2 }

{ n }_{ min }=1.35\xi { g }_{ k }=(1.35\times 0.925\times 6.25)=7.8kN/{ m }^{ 2 }

4. Actions on Sub-Frame C-C

To obtain the forces on subframe C-C, we simply use the tributary width but the load is also increased by 10% in-order to take cognizance of the continuity of the slab.

Maximum Actions

Action\quad from\quad slab\quad ={ e }_{ rf }\times { n }_{ max }\times width\quad =1.1\times 14.55\times 4.8=76.82{ kN/m }

Beam\quad self\quad weight\quad =\quad 5.15kN/m

{ w }_{ max }=\quad 76.82+5.15=82kN/m

Minimum Actions

Action\quad from\quad slab\quad ={ e }_{ rf }\times { n }_{ min }\times width\quad =1.1\times 7.80\times 4.8=41.18{ kN/m }

Beam\quad self\quad weight\quad =\quad 5.15kN/{ m }

{ w }_{ min }=\quad 41.18+5.15=46.3{ kN/m }

The next step will be, to use a suitable method of elastic analysis to analyse the frame for the actions.

Elastic Analysis: Moment Distribution

Now that the actions on the frame have been determined, the moment distribution method will be used to obtain the moment and shears in the beam and connected columns. Figure 5 shows the subframe C-C

Recall, the first thing we do, is to determine the relative stiffness of the inter-connected elements when carrying out moment distribution. Here the relative stiffness is given as I/L. When estimating the stiffness of the flanged beams, the effective flange width is used to be more precise, but we shall consider the beams to be rectangular.

Relative Stiffness

{ k }_{ col,perimeter }=\frac { I }{ l } =\frac { (300\times { 300 }^{ 3 })/12 }{ 3500 } =0.193\times { 10 }^{ 6 }mm^{ 3 }

{ k }_{ col,internal }=\frac { I }{ l } =\frac { (350\times { 350 }^{ 3 })/12 }{ 3500 } =0.357\times { 10 }^{ 6 }mm^{ 3 }

{ k }_{ beam,end }=\frac { I }{ l } =\frac { (300\times { 550 }^{ 3 })/12 }{ 6000 } =0.693\times { 10 }^{ 6 }mm^{ 3 }

{ k }_{ beam,internal }=\frac { I }{ l } =\frac { (300\times { 550 }^{ 3 })/12 }{ 7500 } =0.555\times { 10 }^{ 6 }mm^{ 3 }

Distribution FFactors

Joints 1 & 5

{ D }_{ cols,1 }={ D }_{ cols,5 }=\frac { { k }_{ cols } }{ { k }_{ cols }+{ k }_{ b,ends } } =\frac { (2\times 0.193) }{ (2\times 0.193)+0.693 } =0.36

{ D }_{ 1-2 }={ D }_{ 5-4 }=1-{ D }_{ cols }\quad =1-0.36=0.64

Joints 2 & 4

{ D }_{ cols,2 }={ D }_{ cols,4 }=\frac { { k }_{ cols } }{ { k }_{ cols }+{ k }_{ b,ends }+{ k }_{ b,int } } =\frac { (2\times 0.357) }{ (2\times 0.357)+0.693+0.555 } =0.36

{ D }_{ 2-1 }={ D }_{ 4-5 }=\frac { { k }_{ b.end } }{ { k }_{ cols }+{ k }_{ b,end }+{ k }_{ b,int } } =\frac { 0.693 }{ (2\times 0.357)+0.693+0.555 } =0.35

{ D }_{ 2-3 }={ D }_{ 4-3 }=1-{ D }_{ cols }\quad =1-0.36-0.35=0.29

Joint 3

{ D }_{ cols,3 }=\frac { { k }_{ cols } }{ { k }_{ cols }+2{ k }_{ b,int } } =\frac { (2\times 0.357) }{ (2\times 0.357)+(2\times 0.555) } =0.4

{ D }_{ 3-2 }={ D }_{ 3-4 }=\frac { { k }_{ b.int } }{ { k }_{ cols }+2{ k }_{ b,int } } =\frac { 0.693 }{ (2\times 0.357)+(2\times 0.555) } =0.30

Fixed End Moments

Maximum Actions

{ M’ }_{ 1-2 }=-{ M’ }_{ 2-1 }={ M’ }_{ 4-5 }={ -M }’_{ 5-4 }=\frac { { w }_{ max }{ l }^{ 2 } }{ 12 } =\frac { -82\times { 6 }^{ 2 } }{ 12 } =-246kN.m

{ M’ }_{ 2-3 }=-{ M’ }_{ 3-2 }={ M’ }_{ 3-4 }={ -M }’_{ 4-3 }=\frac { { { w }_{ max }l }^{ 2 } }{ 12 } =\frac { -82\times 7.5^{ 2 } }{ 12 } =-384.4kN.m

Minimum Actions

{ M’ }_{ 1-2 }=-{ M’ }_{ 2-1 }={ M’ }_{ 4-5 }={ -M }’_{ 5-4 }=\frac { { w }_{ min }{ l }^{ 2 } }{ 12 } =\frac { -46.3\times { 6 }^{ 2 } }{ 12 } =-139kN.m

{ M’ }_{ 2-3 }=-{ M’ }_{ 3-2 }={ M’ }_{ 3-4 }={ -M }’_{ 4-3 }=\frac { { { w }_{ min }l }^{ 2 } }{ 12 } =\frac { -46.3\times 7.5^{ 2 } }{ 12 } =–217kN.m\\

Load Cases

As stated earlier in the post, three possible load cases must be considered for us to determine the critical forces in the elements. These are:

Combination-one – All spans are loaded the with maximum load (82kN/m)
Combination-two – Span 1-2 & 3-4 is loaded with the maximum load (82kN/m) while span 2-3 & 4-5 are loaded with the minimum load (46.3kN/m)

Combination Three – Span 2-3 & 4-5 is loaded with the maximum load (82kN/m) while span 1-2 & 3-4 are loaded with the minimum load (46.3kN/m).

Moment Distribution

The moment distribution is carried out same way, we did in the last post. The only concern here, is that the column moments obtained is the sum of the upper and lower column moments. The moment carried by each column will be determined by multiplying by the stiffness’s of the upper and lower columns. In this case the stiffness’s are the same, therefore the moments carried by each columns is simply half of the column moment from the distribution table.


JOINT	1		2			3			4			5
MEMBER	Cols	1-2	2-1	Cols	2-3	3-2	Cols	3-4	4-3	Cols	4-5	5-4	Cols
D.FACTOR	0.36	0.64	0.35	0.36	0.29	0.3	0.4	0.3	0.29	0.36	0.35	0.64	0.36
FEM(kN.m)	-	-246	246	-	-384.4	384.4	-	-384.4	384.4	-	-246	246	-
Distr	88.56	157.44	48.44	49.82	40.14	-	-	-	-40.14	-49.82	-48.44	-157.44	-88.56
CO		24.22	78.72		-	20.07		-20.07	-		-78.72	-24.22
Distr	-8.72	-15.5	-27.55	-28.34	-22.83	-	-	-	22.83	28.34	27.55	15.5	8.72
CO		-13.78	-7.75		-	-11.42		11.42	-		7.75	15.78
Distr	4.96	8.82	2.71	2.79	2.25	-	-	-	-2.25	-2.79	-2.71	-8.82	-4.96
CO		1.36	4.41		-	1.125		-1.125	-		-4.41	-1.36
Distr	-0.49	-0.87	-1.54	-1.59	-1.28	-	-	-	1.28	1.59	1.54	0.87	0.49
CO		-0.77	-0.44			0.64		0.64			0.44	0.77
Distr	0.28	0.49	0.15	0.16	0.13	-	-	-	-0.13	-0.16	-0.15	-0.49	-0.28
M(kN.M)	84.6	84.6	343.2	22.84	-366	393.5	0	393.5	366	-22.84	-343.2	84.6	-84.6


JOINTS	1		2			3			4			5
MEMBERS	Cols	1-2	2-1	Cols	2-3	3-2	Cols	3-4	4-3	Cols	4-5	5-4	Cols
D.FACTOR	0.36	0.64	0.35	0.36	0.29	0.3	0.4	0.3	0.29	0.36	0.35	0.64	0.36
FEM(kN.m)		-246	246		-217	217		-384.4	384.4		-139	139
Distr	88.56	157.44	-10.15	-10.44	-8.41	50.22	66.96	50.22	-71.17	-88.34	-85.9	-88.96	-50.04
C.O		-5.08	78.72		25.11	-4.21		-35.59	25.11		-44.48	-42.95
Distr	1.83	3.25	-36.34	-37.38	-30.11	11.94	15.92	11.94	5.62	6.97	6.78	-27.49	15.46
C.O		-18.17	1.63		5.97	-15.1		2.81	5.97		13.75	3.39
Distr	6.54	11.63	-2.66	-2.73	-2.2	3.69	4.91	3.69	-5.72	-7.1	-6.90	-2.17	-1.22
C.O		-1.33	5.82		1.85	-1.1		-2.86	1.85		-1.09	3.45
Distr	0.48	0.85	-2.68	-2.76	-2.22	1.19	1.58	1.19	-0.22	-0.27	-0.27	2.21	1.24
C.O		-1.34	0.34		0.595	-1.11		-0.11	0.595		1.1	-0.14
Distr	0.48	0.86	-0.36	-0.34	0.29	0.366	0.488	0.366	0.492	-0.61	-0.59	0.09	0.05
M(kN.m)	97.9	-97.9	280.4	-53.7	-226.7	262.9	89.9	-352.7	345.9	-89.4	-256.6	34.5	-34.5


JOINTS	1		2			3			4			5
MEMBERS	Cols	1-2	2-1	Cols	2-3	3-2	Cols	3-4	4-3	Cols	4-5	5-4	Cols
D.FACTOR	0.36	0.64	0.35	0.36	0.29	0.3	0.4	0.3	0.29	0.36	0.35	0.64	0.36
FEM(kN.m)		-139	139		-384.4	384.4		-217	217		-246	246
Distr	50.04	88.96	85.9	88.34	71.17	-50.22	-66.96	-50.22	8.41	10.44	10.15	-157.44	-88.56
C.O		42.95	44.48		-25.11	35.59		4.21	-25.11		-78.72	5.08
Distr	-15.46	27.49	-6.78	-6.97	-5.62	-11.94	-15.92	-11.94	30.11	37.38	36.34	-3.25	1.83
C.O		3.39	-13.75		-5.97	-2.81		15.1	-5.97		-1.63	18.17
Distr	1.22	2.17	6.90	7.1	5.72	-3.69	-4.91	-3.69	2.2	2.73	2.66	-11.63	-6.54
C.O		3.45	1.09		-1.85	2.86		1.1	-1.85		-5.82	1.33
Distr	-1.24	-2.21	0.27	0.27	0.22	-1.19	-1.58	-1.19	2.22	2.76	2.68	0.85	-0.48
C.O		-0.14	-1.1		-0.595	0.11		1.11	-0.595		-0.43	-1.34
Distr	-0.05	0.09	0.59	0.61	0.492	-0.366	-0.488	-0.366	0.29	0.34	0.36	-0.86	-0.482
M(kN.m)	34.5	-34.5	256.6	89.4	-345.9	352.7	-89.9	-262.9	226.7	53.7	-280.4	97.9	-97.9

Bending Moment & Shear-Force Diagrams

Simple rules of static are used to determine the shear forces and bending moment at the span and support from which the diagrams are drawn. The following figures present the B.M.D are presented 6, 7 & 8 for three load cases.

The shear force diagrams have not been included, it should be taken as an exercise.

Finally, when designing these elements, another diagram usually referred to as the moment and shear envelope must be drawn, this contains only the highest forces which would eventually be used for design.

Observation

It is bad practice to analyse and design elements in a frame for the single load case of all span loaded with the maximum loads. For the first load-case of all span loaded with maximum loads, the moments on the column on gridline 3 was equal to zero, this supports the argument that an internal column supporting equal beam lengths should be axially loaded, but we can see that this is not true considering the column developed a significant moment with the second and third load cases. If we were to design this column with the first load-case only, the column will be under-designed. Thus, engineers should cultivate the habit of considering all possible load cases in order to determine the critical forces in the elements.