Structure Analysis I Chapter 4 1 Types of Structures & Loads 1Chapter
Chapter 4 Internal lloading Developed in Structural Members
Internal loading at a specified Point In General The loading for coplanar structure will consist of a normal force N, shear force V, and bending moment M. These loading actually represent the resultants of the stress distribution acting over the member s cross-sectional are
Sign Convention +ve Sign
Procedure for analysis Support Reaction Free-Body Diagram Equation of Equilibrium
Eample 1 Determine the internal shear and moment acting in the cantilever beam shown in figure at sections passing through points C & D
V C M c F y = 0 = 15kN M C = 0 V = 50kN. m C M 5 5 5 = 0 c 5(1) 5(2) 5(3) 20 = 0
kn V V F C D y 20 0 5 5 5 5 0 = = = kn m M M M kn V D D C C. 50 0 20 5(3) 5(2) 5(1) 0 20 = = =
Eample 2 Determine the internal shear and moment acting in section 1 in the beam as shown in figure 18kN RA = RB = 9kN 6kN V F y = 0 V + 9 6 = 0 y = 3kN M at section = 0 M + 6(1) 9(2) = M D = 12kN. m 0
Eample 3 Determine the internal shear and moment acting in the cantilever beam shown in figure at sections passing through points C
V Fy = 0 VC + 9 3 = 0 = 6k M c = 0 M D = 48 k. ft M c + 3(2) 9(6) = 0
Shear and Moment function Procedure for Analysis: 1- Support reaction 2- Shear & Moment Function Specify separate coordinate and associated origins, etending into regions of the beam between concentrated forces and/or couple moments or where there is a discontinuity of distributed loading. Section the beam at distance and from the free body diagram determine V from, M at section
Eample 4 Determine the internal shear and moment Function
Eample 5 Determine the internal shear and moment Function
w 2 15 1 30 2 = = w w 30 2 1 0 30 0 V F = + = 2 1 2 2 ) ( 0.033 30 0 15 30 0 V V F y = + 3 2 1 0.011 30 600 0 600 3 15 ) 30( 0 M M M S + = = + + =
Eample 6 Determine the internal shear and moment Function
0 <12 < 1 1 4 108 0 4 108 0 12 0 V V F y = = + = < < ( ) 2 2 1 1 S 1 2 108 1588 0 4 108 1588 0 4 108 1 M M M V = + + = = 2 1 1 2 108 1588 M + =
12 < < 20 2 V M = F y M 60 S = = 0 = 60 0 2 V M 1300 + 108 48 = + 1588 108 2 0 + 48 ( 6) 2 = 0
Eample 7 Determine the internal shear and moment Function
w 20 9 20 = w w 9 2 1 0 (20) 10 75 0 V F y = + = 2 2 1.11 10 75 9 ) ( V y = ( ) 3 2 3 2 1 2 0.370 5 75 0 9 (20) 10 75 0 M M M S + = = =
Shear and Moment diagram for a Beam
0 ) ( ) ( 0 V V w V F = + + = ( ) O 0 ) ( ) ( 0 ) ( 0 ) ( ) ( 0 M M w M V M w V V V w V F y = + + = = = + + = ε ( ) ( ) 2 O ) ( ) ( ) ( w V M + = ε = = d w V w d dv for ) ( ) ( 0 d V M V dm d ) ( = = d V M V d ) (
Eample 1 Draw shear force and Bending moment Diagram S.F.D B.M.D
Eample 2 Draw shear force and Bending moment Diagram S.F.D B.M.D
Eample 4 18 kn Draw shear force and Bending moment Diagram Ma. moment at = L/2 then wl M = 2 M ma = L 2 wl 8 2 w 2 L 2 2
Eample 3 Eample 3 Draw shear force and Bending moment Diagram
S.F.D B.M.D
Eample 5 Draw shear force and Bending moment Diagram 2 = 14 = 7 M S = M M = 49 14(3.5) + 14(7)
Eample 6a Draw shear force and Bending moment Diagram S.F.D B.M.D
Eample 6b Draw shear force and Bending moment Diagram S.F.D B.M.D
Eample 6c Draw shear force and Bending moment Diagram S.F.D BMD B.M.D
Eample 6d Eample 6d Draw shear force and Bending moment Diagram
Group Work Draw shear force and Bending moment Diagram
Eample 1 Draw shear force and Bending moment Diagram
V(kN)
Eample 2 Draw shear force and Bending moment Diagram
+
Eample 2 Draw shear force and Bending moment Diagram
Eample 3 Draw shear force and Bending moment Diagram
+ + + +
Eample 4 Draw shear force and Bending moment Diagram
+ + +
Problem 1 Draw shear force and Bending moment Diagram
30.5 23.5 + - - + +
Problem 2 Draw shear force and Bending moment Diagram
2 3 at 5 = M M V 5 12 2 3 = 2 = 0 ( R = 11.55 A = 3.46 m ) = 2 3 (3.46)(5)
Eample 1 Draw shear force and Bending moment Diagram Hinge
Reaction Calculation ( ) k A A M y left B 4 0 60 20(5) 10 0 = + = Reaction Calculation k C C M k A y y E y 45 0 60 4(32) 20(27) 5(16) 18(6) (12) 0 4 = = + + + = = k E E E F y 6 0 45 4 20 5 18 0 F 0 0 y = + + + = = = E y 6k =
Frames (Eample 1) Draw Bending moment Diagram
Support reaction & Free Body diagram
S.F.D B.M.D
+ S.F.D - - BMD B.M.D
Frames (Eample 2) Draw shear force and Bending moment Diagram
+ N.F.D + _ S.F.D NFD N.F.D SFD S.F.D BMD B.M.D + - + + B.M.D N.F.D -
Frames (Eample 3) Draw shear force and Bending moment Diagram
N.F.D S.F.D B.M.D - - -
_ N.F.D 64 + 26 S.F.D + B.M.D 251.6
N.F.D S.F.D B.M.D 168
64 + 13.22 S.F.D _ 26 36 _ 31.78 432 _ 432 139.3 _ + 251.6 168+ B.M.D
Frames (Eample 4) Draw shear force and Bending moment Diagram
S.F.D B.M.D + +
_ S.F.D + B.M.D
Frames (Eample 5) Draw shear force and Bending moment Diagram
Frames (Eample 6) Draw shear force and Bending moment Diagram
N.F.D S.F.D B.M.D _
_ N.F.D + _ + S.F.D _ + _ B.M.D
B.M.D S.F.D N.F.D _ + _
Frames (Eample 7) Draw Normal force, shear force and Bending moment Diagram
10kN/m 60kN 20.8 47.7 110 53.7 26.56 o 43.2 26.8 10.5
NFD N.F.D SFD S.F.D BMD B.M.D S.F.D B.M.D
N.F.D S.F.D B.M.D
BMD B.M.D
Moment diagram constructed by the Eample 1 method of superposition
Eample 2.a
Eample 2.b
Problem 1 Draw Normal force, shear force and Bending moment Diagram
Problem 2 Draw Normal force, shear force and Bending moment Diagram