E6002 – Ship Structures I L4 - 1 © C.G.Daley

Lecture 4:

Longitudinal Strength of Ships Murray’s Method for SWBM Estimation Introduction

In this lecture we will • Discuss Murray’s Method to estimate still water bending

moments ~~~~~~~~~~~~~~~~~~~~ Murray’s Method Murray’s method is based on the idea that forces and moments in a ship are self-balancing (no net force or moment is transferred to the world). Any set of weight and buoyancy forces are in balance.

Also, for any cut at x, the moment at the cut can be determined in two ways;

443355

2211)(

LyLyLy

LyLyxBM

−−=

−=

E6002 – Ship Structures I L4 - 2 © C.G.Daley

Murray applied this idea to a ship:

where ff,fa are the distances from the to the centers of weight (fore and aft) gf,ga are the distances from the to the centers of buoyancy (fore and aft) The bending moment at midships is; aaaa gfW ∆−= or ffff gfW ∆−=

These are two ‘estimates’ of the maximum bending moment. We can combine the two, and increase our accuracy, by taking the average of the two;

( ) ( )ffaaffaa ggfWfW ∆+∆−+=21

21

BW BMBM −= weight - buoyancy To find the buoyancy part, Murray suggested

( ) xggBM ffaaB ⋅∆=∆+∆=21

21

where x = average moment arm

E6002 – Ship Structures I L4 - 3 © C.G.Daley

Murray suggested a set of values for x , as a function of the ship length, block coefficient and the ratio of draft to length;

)( bCaLx B +⋅= where T/L a b .03 .209 .03 .04 .199 .041 .05 .189 .052 .06 .179 .063 This table for a and b can be represented adequately by the equation;

LTa /239. −= 003./1.1. −= LTb

Example using Murray’s Method Ship: Tanker L=278m, B=37m, CB=0.8 Assume wave bending moment is; WBMsag = 583800 t-m WBMhog = 520440 t-m The vessel weights, and weight bending moments are as follows;

ITEM Weight lcg Moment(t) (m) (t-m)

Fwd cargo 62000 40 2480000 fuel & water 590 116 68440 steel 12000 55.6 667200

3,215,640 Aft cargo 49800 37 1842600 machinery 3400 125 425000 outfit 900 120 108000 steel 12000 55.6 667200

Σ 140690 t 3,042,800

BMw = 3,129,220

E6002 – Ship Structures I L4 - 4 © C.G.Daley

To find the buoyancy moment we need the draft;

γ⋅⋅⋅⋅=∆= TBLCW B

025.1372788.0140690

⋅⋅⋅=

⋅⋅⋅∆

=γBLC

TB

68.16= m

06.0278

68.16==

LT

Murray’s table gives; a=0.179, b=0.063

32.57)063.8.0179(.278 =+⋅=x m

xBM B ⋅∆=21

428,032,432.5714069021

=⋅= t-m

SWBM = BMW-BMB hog sag 428,032,4220,129,3 −= 145,903−= t-m (- is sag) we need to add the wave bending moment in sag Total BM = 903,145 + 583,800 = 1,486,945 t-m (sag) Note that in this case the ship will never get in the hogging condition, because the SWBM is so large.

E6002 – Ship Structures I L4 - 5 © C.G.Daley

Lecture 4 – Problems. 1. For the example of Murray’s method in the lecture, remove the cargo

weight and add 4000 t of ballast, with a cg of 116m fwd of midship. Re-calculate the maximum sag and hog moments (both still water and wave).

2. For the example of Murray’s method in the lecture, instead of using the weight locations as given, assume that the weights are distributed according to Prohaska. Re-calculate the SWBM.