Top Banner
Province of the EASTERN CAPE EDUCATION DIRECTORATE SENIOR CURRICULUM MANAGEMENT (SEN- FET) HOME SCHOOLING SELF-STUDY WORKSHEET ANSWER SHEET SUBJECT WELDING & METALWORK GRADE DATE APRIL 2020 WK 3 TOPIC FORCES: MEMORANDUM TERM 1 REVISION (Please tick) TERM 2 CONTENT (√) QUESTION 1
8

Curriculumeccurriculum.co.za/Worksheets/Mechanical Technology... · Web viewSELF-STUDY WORKSHEET ANSWER SHEET SUBJECT WELDING & METALWORK GRADE DATE APRIL 2020 WK 3 TOPIC FORCES:

Feb 18, 2021

Download

Documents

dariahiddleston
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript

Province of the

EASTERN CAPE

EDUCATION

DIRECTORATE SENIOR CURRICULUM MANAGEMENT (SEN-FET)

HOME SCHOOLING SELF-STUDY WORKSHEET ANSWER SHEET

SUBJECT

WELDING & METALWORK

GRADE

DATE

APRIL 2020 WK 3

TOPIC

FORCES: MEMORANDUM

TERM 1

REVISION

(Please tick)

TERM 2 CONTENT

(√)

QUESTION 1

QUESTION 2

A diagram below shows a uniform beam. The beam is supported at points RL and RR. Three point loads of 4 N, 5 N and 3 N are exerted onto the beam.

a) Calculate the reactions at supports RL and RR.

b) Calculate the bending moments at point B, C and D.

c) Draw a bending-moment diagram of the beam.

Solution a)

Reactions at the supports RL and RR:

RL x 12 = (3 x 3) + (5 x 6) + (4 x 9)

: RL = 6.25N

RR x 12 = (4 x 3) + (5 x 6) + (3 x 9)

RR = 5.75N

Solution b)

Bending moments:

BMB = (6.25 N x 3 m)

= (18.75 N.m)

BMc = (6.25 x 6) – (4 x 3)

= 25.5 N.m

BMD = (6.25 x 9) – (4 x 6) – (5 x 3)

= 17.25 N.m

Solution c)

Question 3

Solution a)

Reactions at the supports RL and RR

Moments about RR

RL x 12 = (5 x 9) + (6 x 6) + (4 x 3)

RL x 12 = 45 + 36 + 12

RL =

RL = 7,75 N

Reactions about RL

RR x 12 = (4 x 9) + (6 x 6) + (5 x 3)

RR x 12 = 36 + 36 + 15

RR =

RR = 7,25 kN

Solution b)

Shear forces:

𝑆𝐹𝐴 = 0 kN.

𝑆𝐹𝐵 = 7,75 – 5

= 2,75 kN

𝑆𝐹𝐶 = 7,75 – 5 – 6

= - 3, 25 kN

𝑆𝐹𝐷 = 7,75 – 5 – 6 – 4

= - 7,25 kN

𝑆𝐹𝐸 = 7,75 – 5 – 6 – 4 + 7,75

= 0 kN

Solution c)

Bending moments:

Solution d)

Shear force diagram:

Solution e)

Bending moment diagram:

QUESTION 4