Answer for the Original Poster...

I took the case from this - you will see what I've taken as your
situation fromteh numbers I settle on.

formatting link

I note Ray Hayes' explanation, where he looks to have used the same
methodology
Ray Hayes
posted on February 23, 2018, 2:31 pm
# replying to mjacobsen925, Ray Hayes wrote: Your reply is
# foolish. This appears to be a valid question about using a pipe
# section to support a hoist - an overhead lifting design that
# requires calculation. 1.5" black iron pipe is the support beam
# suggested for available electric garage hoists.

# For any who care, stress = M*c/I = M/s. Bending moment = M = WL/2, s
# = 0.326 in^3 for 1.5" sch 40 pipe. Assume yield strength0,000
# psi. L`". stress = M/s = W*L/(2*s). If Safety Factor = 2.0 (low for
# a lifting operation - F.S> should be 6 for overhead lifting):
# Yield/F.S = 30000 psi/2.0 = 15000 psi = W*60/(2*0.326); W = 163
# lbs. Yield Strength (30,000 psi) is the load where the pipe will
# bend without springing back to straight. Original request was for a
# 1" deflection at the center. This is way overloaded - typical beam
# limit might be L/360 = 0.167". But just for yucks: max deflection =
# y = W*L^3/(48

***E***I); E)e6, I = 0.310 in^4 (1.5" sch 40 pipe). y/W =
# 60^3/(48

***29e6&***0.310) = 0.0005, or W/y = 2000, or W = 2000

***y. So to
# get a 1.0" deflection on a 60" pipe (if it did not yield) would be W
# 00 *** 1.0 = 2000 lbs. *BUT the stress with that 2000# load would be
# M/s = W*L/(2

***s) 2000***60/(2

***.326) = 184,000 psi - 6 times the yield
# strength (meaning the pipe would bend to failure). ***For reference,
# limiting deflection to L/360 would allow a load of: W = 2000

***y =
# 2000***(60/360) = 333 lbs. This corresponds to a F.S. = about 1.0.

Dimensions for 1~1/2" Sched 40 tube
Nom OD ID w-thk
1.5 1.900 1.61 0.15

(

*** 1.9 25.4) ;; 48.26
(*** 0.15 25.4) ;; 3.8099999999999996

If I'm not mistaken, that's the dimensions of scaffold tube as we know
it here in the UK.

I can tell you - a 6m length
(- (/ 6e3 25.4) (* 12 19)) ;; 8.220472440944889
19ft 8in length !!!
of scaffold tube supported on its ends will just take my weight in the
middle - about 87kg
(/ 87 0.4536) ;; 191.7989417989418
192lb

I'm going to use a yield strength of 235MPa
There is every likelihood the yield is higher than that.
So the value I'll calculate is the minimum load-bearing possible.

The Young's modulus at 2.1e11Pa is almost independent of steel
strength, so the deflection prediction is invariant of steel grade.

Using my functions
;; moment cap
(* 235e6
(beam-sect-mod-z-d
;; args I, d
(ma2nd-annulus-dia-wthk-cx 48.3e-3 3.8e-3) ; 1.3245821667837055e-07 ; m^4
48.3e-3) ; 5.484812284818656e-06 ; m^3
) ; 1288.9308869323843 ; N.m

So moment capacity is 1289 Newton-metres

(* 60 25.4 1e-3) ;; 1.524 ;; m length

(/
(simple-support-dblbeam-loadcap
;; M_cap, l
1288.9308869323843
1.524) ;; 3383.0207006099326 ;; N (Newtons)
9.81) ; 344.8543017951001 ; kg-f

(/ 345 0.4536) ;; 760.5820105820105
So your curtain rail will bear 760lb at the middle
That's more than enough for a kid swinging on it.

Deflection - which is your question...

(dblsupport-centralload-beam-deflect
;; F(central), l, E, I
(* 344.8543017951001 9.8) ;; 3379.5721575919815 ;; N
1.524 ;; m length of curtain rail
2e11 ;; Elastic modulus of steel (Pa)
(ma2nd-annulus-dia-wthk-cx 48.3e-3 3.8e-3) ;; 1.3245821667837055e-07 ;; m^4 ma-2nd
) ;; 0.00940733235828569 ;; m of deflection at onset of deformation

(/
(

*** 0.00940733235828569 1e3) ;; 9.40733235828569 ;; mm deflection
25.4) ;; 0.370367415680539 **

(* 0.370367415680539 8) ;; 2.962939325444312 ;; about 3/8th-inch

Even at maximum loading of 345kg

* / 760lb the middle of the curtain
rail will be only 9.4mm / *3/8th"

I hope that's the answer to the question.