# Shear force and bending moment for continuous span

It's given that L1 = 5m, L2 and L3= 7.5m, w = 8.9kN/m , I am asked to find the bending moment using these coefficient.. Prior to this , i need to have the value of shear force first.Shear force = coefficient x force.I having problem getting the force for support 2 and 3 .... I think that the 'bottom'shear force for support 2 should be = F x Cs = w x L1 x Cs = 8.9 x 5 x 0.66 = 29.4kN, while 'top'shear force for support 2 should be = F x Cs = w x L1 x Cs = 8.9 x 7.5 x 0.484 = 32.3kN, am i right ? Correct me if i am wrong.

You're on the right track, though there looks to be a typo in the last sentence of your question (you wrote "L1" and then typed the length for L2).

In short, the "bottom" shear coefficient refers to the span extending toward the left. The "top" shear coefficient refers to the span extending toward the right.

Perhaps a brief discussion of how to use coefficient tables/diagrams of this sort might be useful to you. Below is a sketch (not to scale) of a beam configuration, shear diagram, and moment diagram.

For Shear

Conceptually: $$V_{span,\ end} = C_{shear,\ end} * L_{span} * w$$

Taking your figure as an example: $$V_{2,\ right} = 0.516 * L_2 * w$$

Note that the reaction at your Support 3 will equal $$V_{2,\ right} + V_{3,\ left}$$

For Moment

To calculate moment within a span: $$M_{span} = C_{moment} * (L_{span})^2 * w$$

To calculate moment at a support: $$M_{support} = C_{moment} * (L_{average})^2 * w$$ Where $$L_{average}$$ is the average of the spans directly to the left and right of the support under consideration.

Caveat I’m operating on the assumption that these coefficients were provided as part of a homework assignment, along with the beam geometry and loading. It’s important to note that in cases with unequal spans, the exact coefficients will vary with the span ratio(s).

• so, the first span moment should be = 0.058 x ((7.5+5.5)(0.5))^2 x 8.94 = 21.9kNm ? while the first support moment = 0.071 x ((7.5+5.5)(0.5))^2 x 8.94 = 26.81kNm ? – kitzlong Mar 1 '19 at 10:51
• As indicated in the answer @kitzlong, max moment for within a span uses the length of that span only. Moment at a support uses the average length of the two adjacent spans. You might also try checking your work using a 2D beam/frame analysis tool such as FTool. – CableStay Mar 1 '19 at 11:06
• so, the first span moment should be = 0.058 x ((7.5+5.5)(0.5))^2 x 8.94 = 21.9kNm ? while the first support moment = 0.071 x ((5.5))^2 x 8.94 = 19.2kNm ? @CableStay – kitzlong Mar 1 '19 at 11:43
• The moment at fixed support = 0.082 x w x 0.5 L3 , am i right ? – kitzlong Jun 11 '19 at 13:58

Right, or just as a double check, we say:

The reaction at leftmost support plus the lower shear at the 2nd support must add up to total load on the span, meaning (1)*5*8.9kN.

$$0.34 + 0.66 = 1 \quad ,Check.$$

Edit

So,it is 29.37kN at the bottom of the 2nd support and the top shear is

Vtop= 8.9*7.5*0.484=32.307.

• I am aware that 0.34+ 0.66 = 1 , but, that's not my question, I am asking the shear force at support 2.......The 'bottom' shear force for support 2 should be = F x Cs = w x L1 x Cs = 8.9 x 5 x 0.66 = 29.4kN, while 'top' shear force for support 2 should be = F x Cs = w x L1 x Cs = 8.9 x 7.5 x 0.484 = 32.3kN, am i right ? – kitzlong Mar 1 '19 at 5:55
• I am not sure i used the correct F and Li for the Shear force or not .Can someone pls check my working ? For shear force at left of support, i am not sure whether it's F x Cs = w x L1 x Cs = 8.9 x 5 x 0.66 = 29.4kN or F x Cs = w x L2 x Cs = 8.9 x 7.5 x 0.66 = 44.1kN .......Here's my confusion – kitzlong Mar 1 '19 at 5:58
• @kitzlong. Yes, you used the correct F and Li. it is 29.37kN at the bottom. – kamran Mar 1 '19 at 6:26
• @kitzlong, the top part shear for the same point is Vtop= 8.9*7.5*0.484=32.307 – kamran Mar 1 '19 at 6:38