A uniform live load of 1.75 kN/m and a single concentrated live force of 8 kN ar
ID: 1822516 • Letter: A
Question
A uniform live load of 1.75 kN/m and a single concentrated live force of 8 kN are placed on the floor beams. If the beams also support a uniform dead load of 250 N/m, determine (a) the maximum negative shear in panel BC of the girder and (b) the maximum positive moment at B.
A uniform live load of 1.75 kN/m and a single concentrated live force of 8 kN are placed on the floor beams. If the beams also support a uniform dead load of 250 N/m, determine (a) the maximum negative shear in panel BC of the girder and (b) the maximum positive moment at B.Explanation / Answer
let the concentrated live force be 'x' distance from support 'A'.
Using method of super position let's find Girder forces at A, B and C
For super position let's consider the two beams same as given beam, but with different forces.
In beam 1 uniform live load and uniform dead load are acting.
In beam 2 Single concentrated force is acting.
girder forces in given beam = girder forces in beam 1 + girder forces in beam 2
Beam 1
Live load = 1.75 KN/m
Dead load = 250 N/m
so for Girder force at 'A' we consider floor beam between A and B
Net force = 0
Fa + Fb1 = (1750 + 250) * 3 = 6 KN
moment about B gives
Fa * 3 - (1750 + 250) * 3 * (3/2) = 0
Fa = 3 KN
Fb1 = 3 KN ---(1)
for girder force 'B' and 'C' we consider floor beams AB and BD
we know fb1 = 3 KN from equation 1.
for BD floor beam,
net force = 0
fb2 + fc = (1750 + 250) * 3 = 6 KN
moment about C gives,
fb2 * 1.5 - (1750 + 250) * 1.5 * 1.5 /2 + (1750 + 250) * 1.5 * 1.5/2 = 0
fb2 = 0 KN
fc = 6 - fb2 = 6 KN -----(2)
fb = fb1 + fb2 = 3 + 0 = 3 KN
so
fa = 3 KN
fb = 3 KN
fc = 6 KN
FOR BEAM 2
There is a concentrated force of 8 KN acting at 'x' distance from A.
when x <= 3
for floor beam AB,
Net force = 0
fa + fb1 = 8
Moment about A gives
- 8 * x + fb1 * 3 =0
fb1 = 8 *x/3
fa = 8 * ( 1 - x/3)
for floor beam BD,
we have
fb2 + fc = 0
and moment about C gives
fc * 1.5 = 0
so
fc = 0
fb2 = 0
so when
x <= 3
fa = 8 * (1 - x/3)
fb = 8 * x/3
fc = 0
WHEN x > 3
for floor beam AB,
we have
net force = 0
fa + fb1 = 0
and moment about B gives
fa * 1.5 = 0
so
fa = 0
fb1 = 0
for floor beam BD,
we have
Net force =0
fb2 + fc = 8
moment about B gives
- 8 * (x - 3) + fc * 1.5 = 0
fc = 8 * (x - 3)/1.5
fb2 = 8 * ( 1 - x/1.5 + 2) = 8 * ( 3 -x/1.5)
so girder forces when
x > 3
fa = 0
fb = 8 * ( 3 - x/1.5)
fc = 8 * (x -3)/1.5
SO GIRDER FORCES IN GIVEN BEAM
WHEN x <= 3
Fa = fa in beam 1 + fa in beam 2 = 3 + 8 * (1 - x/3)
Fb =fb in beam 1 + fb in beam 2 = 3 + 8 * x/3
Fc = fc in beam 1 + fc in beam 2 = 6 + 0 = 6 KN
WHEN x > 3
Fa = fa in beam 1 + fa in beam 2 = 3 + 0 = 3 KN
Fb =fb in beam 1 + fb in beam 2 = 3 + 8 * ( 3 -x/1.5)
Fc = fc in beam 1 + fc in beam 2 = 6 + 8 * (x-3) /1.5
Shear at point B
when
x <= 3
Moment about point A = 0
Rc * 4.5 - Fb * 3 - Fc * 4.5 =0
Rc = ((3 + 8 * x/3) * 3 + 6 * 4.5)/4.5 = 8 + 8 * x/4.5
Ra = Fa + Fb + Fc - Rc = 3 + 8 + 6 - 8 - 8 * x/4.5 = 9 - 8 * x/4.5
consider beam section AB
let shear and moment at B be vb anf Mb
vb = - (Fa + Fb) + Ra = -(2 + 8 * x/4.5)
Minimum value of shear vb occurs when x = 3
vbmin = - ( 2 +8 * 3/4.5 ) = - 7.33 KN
Moment about point A gives
Mb - Fb * 3 - vb * 3 = 0
Mb = (Fb + vb) *3 = ( 3 + 8 * x/3 - 2 - 8 * x/4.5) = 1 + 8 * x (1/3 - 1/4.5) = 1 + 8 * x * 0.5/4.5
Mb is maximum when x = 3
Mbmax = 1 + 8 * 3 * 0.5 / 4.5 = 3.67 KN.m
When x > 3
Moment about point A = 0
Rc * 4.5 - Fb * 3 - Fc * 4.5 = 0
Rc = (8 * (3 -x/1.5) * 3 + 3 * 3 + 8 * (x -3) /1.5 * 4.5 ) + 6 * 4.5)/4.5 = 8 + 8 * x/4.5
Ra = Fa + Fb + Fc - Rc = 20 - 8 - 8 * x/4.5 = 12 - 8 *x /4.5
consider beam section AB
let shear and moment at B be vb anf Mb
vb = - (Fa + Fb) + Ra = 12 - 8 *x/4.5 - ( 6 + 8 * (3 - x/1.5)) = 6 - 24 + 8 * x (2)/4.5 = 16 * x /4.5 - 18
Vb is minimum when x = 3
vbmin = 16 * 3 /4.5 - 18 = - 7.33 KN
Moment about point A gives
Mb - Fb * 3 - vb * 3 = 0
Mb = (Fb + vb) *3 = ( 3 + 8 * (3 -x/1.5) + 16 * x/4.5 - 18) = 9 - 8 *x/4.5
Mb is maximmum when x = 3
Mbmax = 9 - 8 * 3/4.5 = 3.67 KN.m
So the maximum negative shear in panel BC = - 7.33 KN
So the Maximum Postive moment in at B = 3.67 KN
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