QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

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The snapshots of the program source code and the output is provided below. The source code is provided at the end.

Note:-

1. The main function asks the user to enter the length and width of the plot and passes them as argumnets to the "recommended_litres_fertilizers()" function.

2. The length and width of the plot are passed as arguments to the "recommended_litres_fertilizers()" function. The function calculates the recommended number if litres of fertilizer and returns it to the main function.

Algorithm:-

function recommended_litres_fertilizers(length , width):

area <- length*width;

litres_required <- (area/140);

recommended_litres_required <-(litres_required*1.01400);

return recommended_litres_required;

In case any error is encountered while executing the program , inform in the comments.

Output:-

Source code:-

def recommended_litres_fertilizers(length,width): #length and width of the plot are passed as parameters
    
    total_area = length*width #area of the plot

    fertilizer_required = (total_area/140) # 1 litre of fertilizer if required for 140 metres square of area 

    recommended_amount = fertilizer_required*1.01400
    return recommended_amount

if __name__=="__main__":

    length = int(input("Provide the length of the plot : "))
    width = int(input("Provide the width of the plot : "))

    print("Recommended amount of fertilizer to be used on this plot : {}".format(recommended_litres_fertilizers(length,width)))