KAEA 1121 ENGINEERING MECHANICS

KAEA 1121 ENGINEERING MECHANICSTHREE PINNED ARCHES

An efficient method of connecting two structural members is by inserting a pin or hinge. Examples include the construction of the three pinned arch or the portal frame.

Inserting a pin or hinge to a structure as shown above, changes the structure from being INDETERMINATE to DETERMINATE because a pin or hinge cannot carry moment, i.e moment at the hinge = 0.Any point on the arch or frame is subjected to BENDING MOMENT, SHEAR FORCE and AXIAL FORCE, and these functions can be drawn for each of the structural element. Since a hinge cannot carry any moment, analysis of three pinned structures can be done using the following equations: V = 0, H = 0 and M = 0 ( Equilibrium equations for the whole structure

+ MP = 0 ( Moments taken about the hinge, to the left or right side of the pin

Since the three pinned arch or portal frame is now DETERMINATE (statically), it is easy to analyse.

Example Three pinned arch with concentrated load P

Calculate the support reactions

MA = 0

P . x = VB . L

( VB is known

V = 0

P = VA + VB

( VA is known

H = 0

HA = HB

But MC = 0, Taking moments to the right of the hinge C,

HB x H = VB x L/2 ( ( HB can be found, and hence HAExample 1A three pinned portal frame ABCD is inserted with a hinge at C. If a vertical load of 25 kN acts at B, draw the SF, BM and AF diagrams for this portal. (Hint: The diagrams of these functions consist of straight lines (linear behaviour)

1) Calculate the support reactions

MA = 0, 17 VD = 25 x 5 ( ( VD = 7.35 kN

V = 0, VA + VD = 25 ( ( VA= 17.65 kN

MC = 0, taking moments to the right of C

HD x 5 = VD x 5 ( ( HD = 7.35 kN

H = 0, HA = HD = 7.35 kN2)Bending Moment Calculations

Portal elements are straight and loading is a point load, ( BM diagram is linear.

MA = 0 (Moment at supports = 0)

MB (to the left of member), MB = +VA (5) - HA (5)

= 17.65 (5) (7.35 x 5) = 51.5 kNm (+ve, sag)

MC = 0 ( pinned)

MD = 0 (Moment at supports = 0)

V1 and T1 are the ar and parallel components of force VA

V2 and T2 are the ar and parallel components of force HAV1 and V2 are S.F components, T1 and T2 are A.F components,

Similarly, V3 and V4 are S.F components, T3 and T4 are A.F components,

Shear ForceMember AB, VAB = V1 V2 = 17.65 cos 45o 7.35 cos 45o = 7.28 kN (constant)Member BC, V = 0, (to left of C)

VBC = + VA 25 = - 7.35 kN (constant)

Member CD, VCD (to the right of C)

VCD = V3 V4 = 7.35 cos 45o - 7.35 cos 45o = 0

Axial Force

Member AB, TAB = T1 + T2 = 17.65 sin 45o + 7.35 sin 45o

= -17.68 kN (-ve, compression)

Member BC, H = 0 (to left of C),

TAC = HA = - 7.35 kN (-ve, compression)

Member CD, TCD (to the right of C)

TCD = T3 + T4

= 7.35 sin 45o + 7.35 sin 45o

= - 10.39 kN (-ve, compression)