CALCS PROC012
PIN SHEAR & BENDING 160404

CONTENTS

INTRODUCTION
NOTATION
INPUT TO PROCEDURE
OUTPUT FROM PROCEDURE
PROJECT EXAMPLE INPUT
PROJECT EXAMPLE OUTPUT

INTRODUCTION


Typically for hinges, axles & structural joints. Found in most mechanisms.

SEE ALSO: PROC013

THEORETICAL BASIS: Engineer's Theory of Bending & Roark5 T1 I15.

theory_doc

NOTATION


D diameter-pin, outer
d diameter-pin, inner
R radius, outer
r radius, inner
I 2nd MoA
Py force-transverse
Mz moment-bending
fb stress-bending
fs stress-shear

INPUT TO PROCEDURE


Py & Mz are for a particular Cross section. The user must choose these values but this is advantageous in that loading & support conditions may be reflected in the values used rather than being confined to specific loading states.

OUTPUT FROM PROCEDURE


Shear & Bending stresses.

Output plots use the maximum shear force & bending moment as given in the input data. This loading is depicted as an equivalent loading state for a simply supported beam with central transverse point force. It is understood that the actual loading st ate may differ significantly from the plot equivalent but it is valuable in giving approximate scale to the relationship between maximum moment, constant shear & consequent (equivalent) beam length.

The applied BM & SF indicated on plots are normalised to their respective stresses (fb & fs) as proportions of tension & shear yield stresses. These yield stresses are implicitly scaled to the pin radius and the applied stress are proportioned accord ingly.

PROJECT EXAMPLE INPUT


PROCtitle("MB Ax-G1 Heel TBC"," ","L07")
LET Ngr%=X4:Nk%=AISC:D=20:d=0:Px=0:Py=L07/2:Mx=0:Mz=Py*(20.5/2-13/4):R=D/2:r=d/2



PROJECT EXAMPLE OUTPUT


MB Ax-G1 Heel TBC - PROC012 PIN BEND, SHEAR & TORSION 150325 - L07
A= PI*(R^2-r^2)= 314.2
A= 3.142*(10^2-0^2)= 314.2
I= PI*(R^4-r^4)/4= 7854
I= 3.142*(10^4-0^4)/4= 7855
fa= Px/A= 0
fa= 0/314.2= 0
fb= ABS(Mz)*R/I= 2.177
fb= ABS(1710)*10/7854= 2.177
fs_py= ABS(Py)*.4244*PI*(R^3-r^3)/(4*I*(R-r))= 1.037
fs_py= ABS(244.3)*.4244*3.142*(10^3-0^3)/(4*7854*(10-0))= 1.037
fs_mx= 2*Mx*R/(PI*(R^4-r^4))= 0
fs_mx= 2*0*10/(3.142*(10^4-0^4))= 0
fE= SQR((ABS(fa)+fb)^2+3*fs_mx^2)= 2.177
fE= SQR((ABS(0)+2.177)^2+3*0^2)= 2.177
UF_085= (fa+fb)/(dac(Nk%,Kd%)*mgr(Ngr%,Fy%))= 0.6598
UF_085= (0+2.177)/(0.6*5.5)= 0.6597
UF_086= (fa-fb)/(dac(Nk%,Kd%)*mgr(Ngr%,Fy%))= -0.6598
UF_086= (0-2.177)/(0.6*5.5)= -0.6597
UF_087= (fs_py+fs_mx)/(dac(Nk%,Ks%)*mgr(Ngr%,Fy%))= 0.4713
UF_087= (1.037+0)/(0.4*5.5)= 0.4714
UF_088= fE/(dac(Nk%,Kd%)*mgr(Ngr%,Fy%))= 0.6598
UF_088= 2.177/(0.6*5.5)= 0.6597
END012

PROJECT END